Practice Problems Final

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Most problems have hints if you need to be led through the problem. For some problems, if you get the answer wrong, there is feedback at least giving you the setup.

Problems

Let $S$ be the six outer faces of the box $[0,1] \times [-1,1] \times [-1,0]$ (i.e. $0 \leq x \leq 1$ , $-1 \leq y \leq 1$ , $-1 \leq z \leq 0$ ), oriented with outward normals. Let $\textbf {F} = \left \langle x^2y,xy^2,x^3y^9 \right \rangle$ . Then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {0}.$

The surface $S$ isis not closed. The components of the vector field $\textbf {F}$ have continuous partials everywhere (they are polynomials), so we can use the Divergence Theorem. The surface $S$ is correctlyincorrectly oriented. Now apply the Divergence Theorem.

The setup after using Divergence Theorem is $\int _0^1 \int _{-1}^1 \int _{-1}^0 4xydz\,dy\,dx.$

Switch the order of integration on the integral $\displaystyle \int _0^2 \int _{x^2}^{2\sqrt {2x}} f(x,y)\,dy\,dx$ . Switching the order gives the integral $\int _{\answer {0}}^{\answer {4}} \int _{\answer {y^2/8}}^{\answer {\sqrt {y}}} f(x,y)\,dx\,dy.$

If $\textbf {F} = \left \langle y,x,2z \right \rangle$ and $C$ is the curve given by $\textbf {r}(t) = \left \langle t(t-1)e^{\sqrt {t}},\sin (\pi t^2/2),t/(t^2+1) \right \rangle$ for $0\leq t \leq 1$ , then $\int _{C} \textbf {F} \cdot d\textbf {r} = \answer {1/4}.$

This is a vector line integral, so we first check whether the vector field $\textbf {F}$ is conservative. Notice $\text {curl}(\textbf {F}) = \left \langle \answer {0},\answer {0},\answer {0} \right \rangle ,$ so $\textbf {F}$ is conservative. We can actually find the potential function here, and it is (give the one with $f(0,0,0) = 0$ ) $f(x,y,z) = \answer {xy+z^2}.$ The endpoints of $C$ are $\textbf {r}(1) = (\answer {0},\answer {1},\answer {1/2}),$ and $\textbf {r}(0) = (\answer {0},\answer {0},\answer {0}).$ Plugging these into $f$ and subtracting gives $f(\textbf {r}(1)) - f(\textbf {r}(0)) = \answer {1/4}.$

This is the line integral of a conservative vector field, and $\textbf {F}$ has potential function $f(x,y,z) = xy+z^2$ . Now use FTLI.

If $\textbf {u}$ and $\textbf {v}$ are unit vectors and $\text {proj}_{\textbf {v}}\textbf {u} = \frac {1}{2}\textbf {v}$ , then the angle between $\textbf {u}$ and $\textbf {v}$ is equal to $\answer {\pi /3}$ (give your answer between $0$ and $\pi$ ).

The answer should be $\pi /3$ .

If $\textbf {F} = \left \langle xy,yz,xz \right \rangle$ and $C$ is the triangle in the first octant connecting $(1,0,0)$ , $(0,1,0)$ , and $(0,0,1)$ , oriented clockwise when viewed from above, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {1/2}$ .

Use Stokes’ Theorem.

We first check whether $\textbf {F}$ is conservative. Note that $\text {curl}(\textbf {F}) = \left \langle \answer {-y},\answer {-z},\answer {-x} \right \rangle \neq \left \langle 0,0,0 \right \rangle ,$ so $\textbf {F}$ is not conservative. Doing this line integral directly would involve parametrizing three different line segments, so let us not do that. However, we note that $C$ isis not closed, so we can use Stokes’ theorem. All we need to do is pick a surface $S$ with $C$ as its boundary, and then compute $\displaystyle \iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S}$ instead. One surface we can choose is the portion of the plane through those three points (i.e. filling in the triangle with a plane). This plane has equation $\answer {1}x+\answer {1}y+\answer {1}z = 1.$ Flux integrals depend on orientation, so we decide that now. For Stokes’ theorem to work, we need to pick upwarddownward normals for $S$ .

Now we compute the flux integral. We parametrize $S$ as $\textbf {r}(x,y) = \left \langle x,y,\answer {1-x-y} \right \rangle ,$ with the domain being the triangle $T$ in the $xy$ -plane with vertices $(0,0)$ , $(1,0)$ , and $(0,1)$ . Then $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {1},\answer {1},\answer {1} \right \rangle ,$ which is the wrong normal, so we negate it and use $-(\textbf {r}_x \times \textbf {r}_y)$ instead. The flux is then

$\begin{eqnarray*} \iint_S \text{curl}(\textbf{F}) \cdot d\textbf{S} &=& \iint_T \left\langle \answer{-y},\answer{x+y-1},\answer{-x} \right\rangle \cdot \left\langle -1,-1,-1 \right\rangle dA \\ &=& \iint_T 1dA \\ &=& \text{area}(T) \\ &=& \answer{1/2}. \end{eqnarray*}$

Let $R$ be the region $z \geq x^2+y^2$ and above by $x^2+y^2+z^2 \leq 2$ . Consider the triple integral $\iiint _R zdV.$

In Cartesian coordinates, the bounds on the iterated integral would be $\int _{\answer {-1}}^{\answer {1}} \int _{\answer {-\sqrt {1-x^2}}}^{\answer {\sqrt {1-x^2}}} \int _{\answer {x^2+y^2}}^{\answer {\sqrt {2-x^2-y^2}}} z\,dz\,dy\,dx.$
To get the intersection of the paraboloid $z = x^2+y^2$ with the sphere $x^2+y^2+z^2=2$ , take $x^2+y^2=z$ and plug this into the sphere equation to get $z+z^2=2$ , which gives $z=1$ . The intersection is therefore the circle $x^2+y^2=1$ in the plane $z=1$ .
In cylindrical coordinates, the setup for the integral would be $\int _0^{\answer {2\pi }} \int _{\answer {0}}^{\answer {1}} \int _{\answer {r^2}}^{\answer {\sqrt {2-r^2}}}\answer {zr}dz\,dr\,d\theta .$
Evaluating in either coordinate system gives an answer of $\answer {7\pi /12}$ .

Let $R$ be the solid given by $z \geq \sqrt {x^2+y^2}$ where $1 \leq z \leq 3$ . Evaluate $\displaystyle \iiint _R zdV$ in both cylindrical and spherical coordinates. The answer in either case is $\answer {20\pi }$ .

The setup in cylindrical is either $\int _0^{2\pi } \int _0^1 \int _1^3 zr\,dz\,dr\,d\theta + \int _0^{2\pi } \int _1^3 \int _r^3 zr\,dz\,dr\,d\theta$ or $\int _0^{2\pi } \int _1^3 \int _0^z zr\,dr\,dz\,d\theta .$ In spherical it is $\int _0^{2\pi } \int _0^{\pi /4} \int _{\sec (\phi )}^{3\sec (\phi )} \rho ^3\cos (\phi )\sin (\phi )d\rho \,d\phi \,d\theta .$

The surface area of the surface $S$ given by $\textbf {r}(u,v) = \left \langle 3u+2v,4u+v,\frac {2}{7}v^{7/2} \right \rangle ,$ where $0 \leq u \leq 1$ and $u^{1/4} \leq v \leq 1$ , is equal to $\answer {\frac {2}{3}(2\sqrt {2}-1)}$ .

The surface area is $\displaystyle \iint _S 1dS$ . In this case, this just amounts to calculating $\iint _D \| \textbf {r}_u \times \textbf {r}_v \|dA,$ where $D$ is the domain for $u$ and $v$ . In this case, $\textbf {r}_u = \left \langle \answer {3},\answer {4},\answer {0} \right \rangle ,$ $\textbf {r}_v = \left \langle \answer {2},\answer {1},\answer {v^{5/2}} \right \rangle ,$ $\textbf {r}_u \times \textbf {r}_v = \left \langle \answer {4v^{5/2}},\answer {-3v^{5/2}},\answer {-5} \right \rangle ,$ and $\| \textbf {r}_u \times \textbf {r}_v \| = \answer {5\sqrt {v^5+1}}.$ the integral becomes $\int _0^1 \int _{u^{1/4}}^1 5\sqrt {v^5+1}dv\,du.$ We cannot integrate in this order, so switch the order of integration: $\int _0^1 \int _{\answer {0}}^{\answer {v^4}} 5\sqrt {v^5+1}du\,dv.$

The setup before integrating should be $\int _0^1 \int _0^{v^4} 5\sqrt {v^5+1}du\,dv.$

Let $S$ be the portion of $z = 1-\sqrt {x^2+y^2}$ with $z \geq 0$ , oriented with outward normals, and let $\textbf {F} = \left \langle x^3/3,y^3/3,4 \right \rangle$ . Then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {41\pi /10}.$

Notice that $\text {div}(\textbf {F}) = \answer {x^2+y^2} \neq 0$ , so we cannot use Stokes’ Theorem/change the surface. The surface $S$ is not closed, but we can close it and use Divergence Theorem. Let $S'$ be the disc $x^2+y^2 \leq \answer {1}$ in the plane $z = 0$ , so that $S$ and $S'$ together form a closed surface. We will orient $S'$ upward (arbitrary choice). For Divergence Theorem, $S$ is correctlyincorrectly oriented, and $S'$ is correctlyincorrectly oriented. Therefore, the setup is $\iint _S \textbf {F} \cdot d\textbf {S} - \iint _{S'} \textbf {F} \cdot d\textbf {S} = \iiint _R \text {div}(\textbf {F})dV$ where $R$ is the enclosed solid region.

For the triple integral, it is most nicely computed in coordinates, and the triple integral setup in cylindrical would be $\int _0^{\answer {2\pi }} \int _0^{\answer {1}} \int _{\answer {0}}^{\answer {1-r}} \answer {r^3}dz\,dr\,d\theta .$ This evaluates to $\answer {\pi /10}$ .
For the flux integral, we can parametrize the surface $S'$ as $\textbf {r}(x,y) = \left \langle x,y,\answer {0} \right \rangle$ for $D:\,x^2+y^2 \leq \answer {1}$ . Then $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {0},\answer {0},\answer {1} \right \rangle$ which is the normal. Therefore, the flux of $\textbf {F}$ through $S'$ becomes $\iint _{S'} \textbf {F} \cdot d\textbf {S} = \iint _D \answer {4}dA = 4\text {area}(D) = \answer {4\pi }.$

Putting everything together, we get an answer of $41\pi /10$ .

Let $\textbf {F} = \left \langle \sin (yz^5),x,\arctan (xyz^2) \right \rangle$ and let $S$ be the portion of $z = 1-x^2-y^2$ with $z \geq 0$ , oriented with outward normals. Then $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \answer {\pi }.$

There are two ways to do this: surface independence/divergence theorem (with closing) or direct Stokes’ Theorem. The latter would say $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \int _C \textbf {F} \cdot d\textbf {r},$ where $C$ is the boundary of $S$ . The boundary is $x^2+y^2 = \answer {1}$ in the plane $z = \answer {0}$ , oriented clockwisecounterclockwise when viewed from above. Parametrize the circle as $\textbf {r}(t) = \left \langle \cos (t),\sin (t),\answer {0} \right \rangle ,$ for $0 \leq t \leq 2\pi$ , which is oriented correctlyincorrectly . The line integral of $\textbf {F}$ over $C$ is therefore $\int _0^{2\pi } \textbf {F}(\textbf {r}(t)) \cdot \textbf {r}'(t)dt = \int _0^{2\pi } \answer {\cos ^2(t)}dt = \answer {\pi }.$ You should try using surface independence/divergence theorem (with closing) as well.

Let $P$ be the plane through the point $(1,2,0)$ which is perpendicular to the line given by $x = 3; \quad \frac {y-1}{2} = \frac {z+1}{5}.$ An equation for $P$ is (give the answer in the form $ax+by+cz=d$ ): $\answer {0}x+\answer {2}y+\answer {5}z = 4.$

A normal vector for the plane is the direction vector for the line, which is $\left \langle 0,2,5\right \rangle$ . The answer is $2y+5z=4$ .

If $\textbf {F} = \left \langle x,y,z \right \rangle$ and $S$ is the portion of $x^2+y^2+z^2=1$ where $z \geq 0$ , oriented with outward normals, then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {2\pi }.$

Note that $S$ is not closed, but we can close it and try to use Divergence Theorem. Note that $\text {div}(\textbf {F}) = \answer {3}$ . We can close it by adding $S'$ which will be the disc $x^2+y^2 \leq \answer {1}$ in the plane $z = \answer {0}$ . We will orient $S'$ downward (arbitrary choice). For Divergence Thoerem, the surface $S$ is correctlyincorrectly oriented and $S'$ is correctlyincorrectly oriented. Therefore, $\iint _S \textbf {F} \cdot d\textbf {S} + \iint _{S'} \textbf {F} \cdot d\textbf {S} = \iiint _R \text {div}(\textbf {F})dV$ where $R$ is the enclosed solid.

The triple integral is $\iiint _R 3dV =3 \text {vol}(R) = \answer {2\pi }.$
The flux through $S'$ : Parametrize the disc as $\textbf {r}(x,y) = \left \langle x,y,\answer {0} \right \rangle$ for $D:\,x^2+y^2 \leq 1$ . Then $\textbf {r}_x \times \textbf {r}_y = \left \langle 0,0,\answer {1} \right \rangle ,$ which is the normal, so we switch the sign on the vector to get a normal of $\left \langle 0,0,-1 \right \rangle$ . Using this we get a flux of $\answer {0}$ .

Therefore, the answer is $\answer {2\pi }$ .

Close and use Divergence Theorem, or just compute the flux directly. You should get an answer of $2\pi$ .

If $R$ is the disc $x^2+(y-1)^2 \leq 1$ , then $\displaystyle \iint _R ydA = \answer {\pi }$ .

The setup should be $\int _0^{\pi } \int _0^{2\sin (\theta )} r^2\sin (\theta )dr\,d\theta .$

If $\textbf {F} = \left \langle x/4,y/9 \right \rangle$ , evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ , where $C$ is the curve $x^2/4+y^2/9 = 1$ , in three different ways. No matter which way you do it, you get an answer of $\answer {0}$ .

You can use the fact that $\textbf {F}$ is conservative, you can use Green’s theorem, or you can do the line integral directly.

Let $R$ be the solid region $x^2+y^2+z^2 \leq 1$ where $y \leq x$ , and let $S$ be the outer facing of the solid, oriented with outward normals (i.e. $S$ is the closed surface consisting of the hemisphere and the disc to close the hemisphere). If $\textbf {F} = \left \langle x^3-3x,y^3+xy,z^3-xz\right \rangle$ , then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {-4\pi /5}.$

Since $S$ is closed, we can use Divergence Theorem. The surface $S$ is correctly oriented. The divergence of $\textbf {F}$ is $\text {div}(\textbf {F}) = \answer {3x^2+3y^2+3z^2-3}$ . We can set the triple integral up in spherical. It becomes $\int _{-3\pi /4}^{\pi /4} \int _0^{\pi } \int _0^1 (3\rho ^2-3)\rho ^2\sin (\phi )d\rho \,d\phi \,d\theta .$

Let $C$ be the parallelogram with vertices $(0,0,0)$ , $(1,0,1)$ , $(1,3,1)$ , and $(0,3,0)$ , oriented clockwise when viewed from above. If $\textbf {F} = \left \langle \sin (x^2)+y^2,\cos (y^2)+x+z,e^{z^2}+\cos (x^{10}) \right \rangle ,$ then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {3}.$

Notice $\text {curl}(\textbf {F}) = \left \langle \answer {-1},\answer {10x^{9}\sin (x^{10})},\answer {1-2y} \right \rangle ,$ which tells us that $\textbf {F}$ isis not conservative. To do this integral directly would involve four different line segments, so we would like to avoid that. For Stokes’ Theorem, The curl isn’t particularly nice, but maybe something nice will happen when we take the flux. At this point, it seems like the only option.

The curve $C$ lies on a plane, and the equation of the plane is $z = \answer {x}$ . We will let this be our surface $S$ . We can parametrize the surface as $\textbf {r}(x,y) = \left \langle x,y,\answer {x} \right \rangle ,$ and the domain $D$ is a rectangle given by $\answer {0} \leq x \leq \answer {1},$ $\answer {0} \leq y \leq \answer {3}.$ With the given orientation of $C$ , the surface $S$ should be oriented with normals pointing in the positivenegative $z$ -direction. Now we compute $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {-1},\answer {0},\answer {1} \right \rangle ,$ which is the correctincorrect normal vector, so we can negate it when computing the flux. The flux of the curl is therefore

$\begin{eqnarray*} \iint_S \text{curl}(\textbf{F}) \cdot d\textbf{S} &=& \iint_D (\text{curl}(\textbf{F}))(\textbf{r}(x,y)) \cdot -(\textbf{r}_x \times \textbf{r}_y)\,dA \\ &=& \iint_D \left\langle -1,10x^9\sin(x^{10}),1-2y \right\rangle \cdot \left\langle \answer{1},\answer{0},\answer{-1} \right\rangle\,dA \\ &=& \iint_D \answer{-2+2y} \,dA \\ &=& \int_0^1 \int_0^3 (-2+2y)dy\,dx \\ &=& \answer{3}. \end{eqnarray*}$

The answer should be $3$ .

If $\textbf {F} = \left \langle 0,0,xz \right \rangle$ and $C$ is the curve given by $\textbf {r}(t) = \left \langle \cos (t),\sin (t),t \right \rangle$ for $t \in [0,2\pi ]$ , then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {0}$ .

Doing this integral directly is the best way to go since $\textbf {F}$ is not conservative and $C$ is not closed (and this is a 3D vector field, not 2D where you can close and use Green’s Theorem).

Doing this directly, the setup is $\int _0^{2\pi } t\cos (t)dt.$ Use integration by parts.

Let $S$ be the sphere $x^2+y^2+z^2=1$ , oriented with outward normals, and let $\textbf {F} = \left \langle zy^4-y^2,y-x^3,z^2 \right \rangle$ . The integral $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \answer {0}.$

If $\textbf {F} = \left \langle x,y,1-2z \right \rangle$ and $S$ is the portion of $x^2+y^2+z^2 = 4$ where $z \leq 1$ , oriented with outward normals, then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {3\pi }$ .

This is either a Divergence Theorem after closing the surface or just straight changing the surface problem, since $\text {div}(\textbf {F}) = 0$ . Let $S'$ be the disc $x^2+y^2 \leq 3$ in the plane $z=1$ , oriented downward. Then the flux of $\textbf {F}$ through $S$ is the same as the flux through $S'$ . Calculate the flux through $S'$ .

If $xy+xz+yz+x^2y^2z^2=4$ , then $\frac {\partial z}{\partial x}\Big |_{(x,y,z) = (1,1,1)} = \answer {-1}.$

This is an implicit differentiation problem. Let $F(x,y,z) = xy+xz+yz+x^2y^2z^2$ . Near $(1,1,1)$ , $z$ becomes a function of $x$ and $y$ , so the equation reads $F(x,y,z(x,y)) = 4$ . Taking the partial of both sides with respect to $x$ and using the chain rule on the left side gives $F_x + F_z\frac {\partial z}{\partial x} = 0,$ so $\frac {\partial z}{\partial x} = -\frac {F_x}{F_z} = -\frac {\answer {y+z+2xy^2z^2}}{\answer {x+z+2yx^2z^2}}.$ Plugging in the point $(1,1,1)$ gives an answer of $-1$ .

The triple integral $\int _0^3 \int _0^{\sqrt {9-x^2}} \int _0^{\sqrt {x^2+y^2}} y dz\,dy\,dx$ written as a triple integral in spherical coordinates would be (write “rho” for $\rho$ , $p$ for $\phi$ , and “theta” for $\theta$ ): $\int _0^{\answer {\pi /2}} \int _{\answer {\pi /4}}^{\answer {\pi /2}} \int _{\answer {0}}^{\answer {3\csc (p)}} \answer {\rho ^3 \sin ^2(p)\sin (\theta )} d\rho \,d\phi \,d\theta .$

If $S$ is the portion of $x+2y+z=1$ in the first octant, then $\displaystyle \iint _S ydS = \answer {\sqrt {6}/24}$ .

This is a scalar surface integral. Parametrize $S$ as $\textbf {r}(x,y) = \left \langle x,y,\answer {1-x-2y} \right \rangle$ with domain the filled-in triangle in the $xy$ -plane with vertices $(0,0)$ , $(1,0)$ and $(0,1/2)$ , call this $T$ Then $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {1},\answer {2},\answer {1} \right \rangle$ which has magnitude $\answer {\sqrt {6}}$ . The surface integral is therefore $\sqrt {6}\iint _T ydA = \sqrt {6}\int _0^{1/2} \int _0^{\answer {1-2y}} y\,dx\,dy.$ Now evaluate.

The answer is $\sqrt {6}/24$ . See hint for steps.

The double integral $\int _0^2 \int _x^2 e^{y^2}dy\,dx = \answer {\frac {1}{2}(e^4-1)}.$

Switch the order of integration.

Switch the order of the double integral to $\int _0^2 \int _0^y e^{y^2}dy\,dx.$

The surface area of the part of $z = x^2+y^2$ below $z=9$ is $\answer {\frac {\pi }{6}(37^{3/2}-1)}$ .

Let $S$ be the portion of $z = x^2+y^2$ below $z=9$ . We need to calculate $\iint _S 1dS.$ We need to parametrize $S$ . We can parametrize $S$ as $\textbf {r} = \left \langle x,y,\answer {x^2+y^2} \right \rangle$ with domain $D:\,x^2+y^2 \leq \answer {9}$ . The magnitude $\| \textbf {r}_x \times \textbf {r}_y \| = \sqrt {\answer {1+4x^2+4y^2}}.$ The surface integral is therefore $\iint _D \sqrt {1+4x^2+4y^2}\,dA.$ We can go to polar from here: $\int _0^{\answer {2\pi }} \int _0^{\answer {3}} \answer {r\sqrt {1+4r^2}}\,dr\,d\theta .$

The setup should be $\int _0^{2\pi } \int _0^3 r\sqrt {1+4r^2}\,dr\,d\theta .$

If $\textbf {F} = \left \langle 5x^3,5y^3,5z^3 \right \rangle$ and $S$ is the sphere $x^2+y^2+z^2=2$ , oriented with outward normals, then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {48\pi \sqrt {2}}.$

Use the Divergence Theorem.

Using Divergence Theorem and then transforming the triple integral into spherical coordinates, we get $\int _0^{2\pi } \int _0^{\pi } \int _0^{\sqrt {2}} 15\rho ^4 \sin (\phi ) d\rho \,d\phi \,d\theta .$

Let $\textbf {F} = \left \langle x-y+\sin (z^2y),x+y+\tan ^{-1}(xz),z+z^2 \right \rangle$ . Let $S$ be the surface which is the portion of $z = 4-x^2-y^2$ above the $xy$ -plane, oriented with inward normals. Then $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \answer {-8\pi }.$

You can either use Stokes’ Theorem or surface independence. Using the former, we can say $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \int _C \textbf {F} \cdot d\textbf {r},$ where $C$ is the boundary of $S$ (with positive boundary orientation). In this case, the boundary is the circle $x^2+y^2 = \answer {4}$ in the plane $z=\answer {0}$ , oriented clockwisecounterclockwise when viewed from above. We can parametrize the boundary as $\textbf {r}(t) = \left \langle 2\cos (t),2\sin (t),0 \right \rangle$ for $0 \leq t \leq 2\pi$ . This is going the correctwrong way, so we add a negative. Therefore, the answer is $-\int _0^{2\pi } \textbf {F}(\textbf {r}(t)) \cdot \textbf {r}'(t)dt.$

The answer is $-8\pi$ .

Let $S$ be the portion of $x^2+y^2+z^2=2$ above the plane $z=1$ , oriented with downward normals. Let $\textbf {F} = \left \langle z-y,0,y \right \rangle$ . Then $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \answer {-\pi }.$ Try to get this answer in two different ways.

Notice that $\text {curl}(\textbf {F}) = \left \langle 1,1,1 \right \rangle$ . We can use surface independence to get the flux of this vector field. Let $S'$ be the disc $x^2+y^2 \leq 1$ in the plane $z = \answer {1}$ , oriented with downward normals. Then $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \iint _{S'} \text {curl}(\textbf {F}) \cdot d\textbf {S}.$ Now calculate the flux through $S'$ .

Alternatively, you can use Stokes’ Theorem.

The answer is $-\pi$ .

Let $R$ be the region given by $0 \leq z \leq 1-\sqrt {x^2+y^2}$ . The integral $\iiint _R z\,dV = \answer {\pi /12}.$

This is best done in cylindrical, and the setup should be $\int _0^{2\pi } \int _0^1 \int _0^{1-r} zr\,dz\,dr\,d\theta .$

If $R$ is the solid region bounded by $x = y^2$ , $z=x$ , $z=0$ , and $x=1$ , then the volume of $R$ is equal to $\answer {4/5}$ .

One possible setup is $\int _0^1 \int _0^x \int _{-\sqrt {x}}^{\sqrt {x}} 1\,dy\,dz\,dx.$

The volume of the solid given by $z \geq x^2+y^2$ and $z \leq 36 - 3x^2-3y^2$ is equal to $\answer {162\pi }$ .

This is good in cylindrical coordinates: $\int _0^{2\pi } \int _0^3 \int _{r^2}^{36-3r^2} r\,dz\,dr\,d\theta .$

Let $R$ be the solid given by $\sqrt {x^2+y^2} \leq z \leq \sqrt {3x^2+3y^2}$ and $x^2+y^2+z^2 \leq 1$ . The integral setup for $\displaystyle \iiint _R z\,dV$ in spherical coordinates is (input “rho” for $\rho$ , “p” for $\phi$ ): $\int _0^{\answer {2\pi }} \int _{\answer {\pi /6}}^{\answer {\pi /4}} \int _{\answer {0}}^{\answer {1}} \answer {\rho ^3\cos (p)\sin (p)}d\rho \,d\phi \,d\theta .$

The answer is $\int _0^{2\pi } \int _{\pi /6}^{\pi /4} \int _0^1 \rho ^3\cos (\phi )\sin (\phi )d\rho \,d\phi \,d\theta .$

Let $R$ be the solid region above $z = x^2+y^2$ and below $z = 2y$ . The triple integral setup for $\displaystyle \iiint _R zdV$ in cylindrical coordinates is $\int _0^{\answer {\pi }} \int _{\answer {0}}^{\answer {2\sin (\theta )}} \int _{\answer {r^2}}^{\answer {2r\sin (\theta )}} \answer {zr}\,dz\,dr\,d\theta .$

The projection into the $xy$ -plane of the intersection of the paraboloid and the plane is a circle of radius $1$ and center $(0,1)$ , which has polar equation $r = 2\sin (\theta )$ . The setup should be $\int _0^{\pi } \int _0^{2\sin (\theta )} \int _{r^2}^{2r\sin (\theta )} zr\,dz\,dr\,d\theta .$

The tangent plane to the surface $z = 2x^2+3y^2$ at the point $(1,1,5)$ has equation (give in the form $ax+by+cz=d$ ): $\answer {4}x+\answer {6}y-z = \answer {5}.$

A normal vector for the plane is $\left \langle f_x(1,1),f_y(1,1),-1 \right \rangle$ , where $f(x,y) = 2x^2+3y^2$ . The answer is $4x+6y-z=5.$

Let $L$ be the line of intersection between the planes $2x+3y=5$ and $x+3y+z=1$ .

The point $(1,1,2)$ on the line $L$ .
The vector $\left \langle 3,-2,3 \right \rangle$ parallel to the line $L$ .

The area of the triangle with vertices $A(1,2,1)$ , $B(2,3,0)$ , and $C(1,0,1)$ is equal to $\answer {\sqrt {2}}$ .

The area is equal to, for example, $\frac {1}{2} \| \overrightarrow {AB} \times \overrightarrow {AC} \|.$ The vectors $\overrightarrow {AB} = \left \langle \answer {1},\answer {1},\answer {-1} \right \rangle ,$ $\overrightarrow {AC} = \left \langle \answer {0},\answer {-2},\answer {0} \right \rangle ,$ which have a cross product of $\overrightarrow {AB} \times \overrightarrow {AC} = \left \langle \answer {-2},\answer {0},\answer {-2} \right \rangle .$ This has magnitude $\answer {\sqrt {8}}$ , giving an area of $\sqrt {2}$ .

Let $S$ be the surface given implicitly by $2x^2+3y^2-5xyz+3z^3 = 3$ . The tangent plane to $S$ at the point $P(1,1,1)$ has equation (give the answer in the form $ax+by+cz=d$ ): $\answer {-1}x+\answer {1}y+\answer {4}z = 4.$

Let $F(x,y,z) = 2x^2+3y^2-5xyz+3z^3$ . Then the tangent plane to $F(x,y,z) = 3$ at the point $P$ has a normal vector of $\nabla F(P)$ . Here, the gradient is $\nabla F(x,y,z) = \left \langle \answer {4x-5yz},\answer {6y-5xz},\answer {-5xy+9z^2} \right \rangle ,$ so $\nabla F(P) = \left \langle -1,1,4 \right \rangle .$ Since we have a point $P$ on the plane, we can write down the equation as $-(x-1)+(y-1)+4(z-1) = 0.$ Now rearrange to get the equation of the plane in $ax+by+cz=d$ form.

Let $f(x,y) = 2xy+y^2$ and $P$ be the point $P(1,-2)$ . The instantaneous rate of change of the function $z=f(x,y)$ at $P$ in the direction of $Q(5,1)$ is equal to $\answer {-22/5}$ .

Is $z = f(x,y)$ increasing or decreasing at $P$ in the direction of $Q$ ?

Increasing Decreasing

If $xy+xz+yz+x^2y^2z^2=4$ , then $\frac {\partial z}{\partial x}\Big |_{(x,y,z) = (1,1,1)} = \answer {-1}.$

Suppose $g(u,v) = (u+v,u^2v,u+3v)$ and $f(x,y,z) = x+y^2+2z+xyz$ , and let $h(u,v) = f(g(u,v))$ . Then $\frac {\partial h}{\partial u}\Big |_{(u,v) = (2,-1)} = \answer {39}$

This is a chain rule problem. We get $\frac {\partial h}{\partial u} = \frac {\partial f}{\partial x}\frac {\partial x}{\partial u} +\frac {\partial f}{\partial y}\frac {\partial y}{\partial u} +\frac {\partial f}{\partial z}\frac {\partial z}{\partial u}.$ Notice $\frac {\partial x}{\partial u} = \answer {1},\quad \frac {\partial y}{\partial u} = \answer {2uv},\quad \frac {\partial z}{\partial u} = \answer {1},$ $\frac {\partial f}{\partial x} = \answer {1+yz}, \quad \frac {\partial f}{\partial y} = \answer {2y+xz}, \quad \frac {\partial f}{\partial z} = \answer {2+xy}.$ Finally, notice that the point $(u,v) = (2,-1)$ corresponds to an $(x,y,z)$ value of $(x,y,z) = (\answer {1},\answer {-4},\answer {-1})$ (this is $g(2,-1)$ ). Plugging all into the above expression gives an answer of $39$ .

Let $R$ be the region under the plane $z=5$ , above the paraboloid $z = 1-x^2-y^2$ , and inside the cylinder $x^2+y^2=1$ . The volume of $R$ is equal to $\answer {9\pi /2}$ .

The answer is $\int _0^{2\pi } \int _0^1 \int _{1-r^2}^5 r\,dz\,dr\,d\theta .$

Let $S$ be the portion of the cylinder $x^2+y^2=4$ with $0 \leq z \leq 1$ , oriented with outward normals. If $\textbf {F} = \left \langle x^3+z^{10},y^3+\sin (xz),z^2 \right \rangle$ , then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {24\pi }.$

We can close this and use Divergence Theorem, but we have to close it with two different lids: $x^2+y^2 \leq 4$ in the plane $z=0$ and in the plane $z=1$ . Let $S'$ be the disc $x^2+y^2 \leq 4$ in the plane $z=0$ , and $S''$ be the disc $x^2+y^2 \leq 4$ in the plane $z=1$ , both oriented upward. Then we can use Divergence Theorem on combined surface $S$ with $S'$ and $S''$ , noting that $S$ is correctlyincorrectly oriented, $S'$ is correctlyincorrectly oriented, and $S''$ is correctlyincorrectly oriented. Therefore $\iint _S \textbf {F} \cdot d\textbf {S} - \iint _{S'} \textbf {F} \cdot d\textbf {S} + \iint _{S''} \textbf {F} \cdot d\textbf {S} = \iiint _R \text {div}(\textbf {F})\,dV.$ The triple integral should give an answer of $28\pi$ , the flux through $S'$ is $0$ and the flux through $S''$ is $4\pi$ . Together, this gives an answer of $24\pi$ .