Practice Problems Midterm 2

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The volume of the solid which lies between $z = x\sec ^2(y)$ and $z=0$ and above the region in the $xy$ -plane bounded by $x=0$ , $x=2$ , $y =0$ , and $y = \pi /4$ is equal to $\answer {2}$ .

The setup should be $\int _0^{\pi /4} \int _0^2 x\sec ^2(y)dx\,dy.$

If $R = [-1,1] \times [0,1]$ , and $f(x,y) = \frac {xy}{1+x^2}$ , then $\iint _R f(x,y)dA = \answer {0}.$

If $R = [-\pi ,\pi ] \times [-\pi ,\pi ]$ , then $\iint _R (1+x^2\sin (y)+y^2\sin (x))dA = \answer {4\pi ^2}.$

If $T$ is the triangle in the $xy$ -plane with vertices $(0,0)$ , $(1,2)$ , and $(2,2)$ , then the surface area of the portion of the surface $z = 7+\sqrt {8}\,x+y^2$ lying above $T$ is $\answer {49/12}$ .

Let $D$ be the region in the $xy$ -plane bounded by $y = x-1$ and $y^2 =2x+6$ . Consider the integral $\displaystyle \iint _D xydA$ .

This double integral in the order $dx\,dy$ is equal to $\int _{y=\answer {-2}}^{y=\answer {4}} \int _{x = \answer {\frac {1}{2}y^2-3}}^{x = \answer {y+1}} xydx\,dy.$
This double integral in the order $dy\,dx$ is equal to $\int _{x=-3}^{x=\answer {-1}} \int _{y = \answer {-\sqrt {2x+6}}}^{y = \answer {\sqrt {2x+6}}} xydy\,dx+\int _{x=\answer {-1}}^{x=5} \int _{y = \answer {x-1}}^{y = \answer {\sqrt {2x+6}}} xydy\,dx.$
The value of the double integral is $\answer {36}$ .

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

Switch the order of integration to get $\int _0^1 \int _0^y \sin (y^2)dx\,dy.$

If $D$ is the region in the $xy$ -plane bounded by $y=4$ , $y=x$ and $x=0$ , then $\iint _D y^2e^{xy}dA = \answer {\frac {1}{2}(e^{16}-17)}.$

The setup is $\int _0^4 \int _x^4 y^2e^{xy}dy\,dx,$ or $\int _0^4 \int _0^y y^2e^{xy}dx\,dy.$

The double integral $\int _0^4 \int _{\sqrt {y}}^2 e^{x^3}dx\,dy = \answer {\frac {1}{3}(e^8-1)}.$

Switch the order of integration to get $\int _0^2 \int _0^{x^2} e^{x^3}dy\,dx.$

If $D$ is the region in the $xy$ -plane given by $D:\, 0 \leq y \leq 1-|x|$ , then $\iint _D (2+x)dA = \answer {2}.$

If $D$ is the filled-in triangle with vertices $(0,1)$ , $(1,2)$ , and $(4,1)$ , then $\iint _D y^2dA = \answer {11/3}.$

The setup is $\int _1^2 \int _{y-1}^{7-3y} y^2dx\,dy.$

Consider the vector field $\textbf {G} = \left \langle x,y,z \right \rangle$ . The divergence of $\textbf {G}$ is $\text {div}(\textbf {G}) = \answer {3}.$

Is there a vector field $\textbf {F}$ with $\textbf {G} = \text {curl}(\textbf {F})$ ? YesNo .

The volume of the solid bounded by $y=x^2$ and the planes $z=0$ and $y+z=1$ is $\answer {8/15}$ .

The setup is $\int _{-1}^1 \int _{x^2}^1 (1-y)dy\,dx.$

If $D$ is the region in the first quadrant given by $y\geq 0$ , $y \leq \sqrt {3}\,x$ and $x^2+y^2\leq 9$ , then $\iint _D (x^2+y^2)^{3/2}dA = \answer {81\pi /5}.$

The setup in polar is $\int _0^{\pi /3} \int _0^{3} r^4dr\,d\theta .$

The volume of the solid that lies under $z=x^2+y^2$ and above the $xy$ -plane, inside the cylinder $x^2+y^2=2x$ is equal to $\answer {3\pi /2}$ .

The setup in polar is $\int _{-\pi /2}^{\pi /2} \int _0^{2\cos (\theta )} r^3dr\,d\theta .$

Decide whether the following expressions make sense for scalar functions $f$ , $g$ , and vector fields $\textbf {F}$ and $\textbf {G}$ .

$\text {curl}(\textbf {F} \times \nabla f)$ . Makes sense.Doesn’t make sense.
$\text {div}(\textbf {F} \cdot \textbf {G})$ . Makes sense.Doesn’t make sense.
$\text {div}(\nabla (fg))$ . Makes sense.Doesn’t make sense.
$\text {curl}(\text {div}(\textbf {F}))$ . Makes sense.Doesn’t make sense.
$\text {div}(\textbf {F}) + \nabla f$ . Makes sense.Doesn’t make sense.

The area of the portion of $x^2+y^2\leq 2$ where $x \geq 1$ and $y \geq 0$ is $\answer {\pi /4 - 1/2}$ .

The setup in polar is $\int _0^{\pi /4} \int _{\sec (x)}^{\sqrt {2}} rdr\,d\theta .$

If $C$ is the arc of the parabola $y = x^3$ from $(0,0)$ to $(1,1)$ , then $\int _C x^3ds = \answer {\frac {1}{54}(10\sqrt {10}-1)}.$

Parametrize the curve as $\textbf {r}(t) = \left \langle t,t^3 \right \rangle$ for $0 \leq t \leq 1$ . Then $\int _C x^3ds =\int _0^1 t^3 \sqrt {1+9t^4}dt.$

Consider the vector field $\textbf {F} = \left \langle x\sin (y),0,xyz \right \rangle$ .

The vector field $\textbf {F}$ isis not conservative.
The divergence $\text {div}(\textbf {F}) = \answer {\sin (y)+xy}$ .
If $C$ is the curve given by $\textbf {r}(t) = \left \langle t,t^2,t^3 \right \rangle$ for $0 \leq t \leq 1$ , then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {\frac {5}{6}-\frac {1}{2}\cos (1)}.$

The surface area of $z = \sqrt {x^2+y^2}$ for $0 \leq z \leq 1$ is $\answer {\pi \sqrt {2}}$ .

Parametrize the surface as $\textbf {r}(r,\theta ) = \left \langle r\cos (\theta ),r\sin (\theta ),r \right \rangle$ for $0 \leq r \leq 1$ , $0 \leq \theta \leq 2\pi$ . Then use the formula $\int _0^{2\pi } \int _0^1 \| \textbf {r}_r \times \textbf {r}_{\theta } \| dr\,d\theta .$

Suppose $\textbf {F} = \left \langle ay+x,bx^2+x+1 \right \rangle$ is a vector field, where $a,b$ are constants.

The values of $a$ and $b$ that make $\textbf {F}$ a conservative vector field are $a = \answer {1},\quad b = \answer {0}.$
Let $C$ be the curve given by $\textbf {r}(t) = \left \langle t\cos (\pi t),\sin (2\pi t) \right \rangle$ for $0\leq t \leq 2$ . For the values of $a$ and $b$ you found in the previous part, the integral $\int _C \textbf {F} \cdot d\textbf {r} = \answer {2}.$

Let $C$ be the portion of $y=x^2$ from $(-1,1)$ to $(1,1)$ followed by the line segment from $(1,1)$ to $(-1,1)$ . If $\textbf {F} = \left \langle y^2,4xy \right \rangle$ , then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {8/5}.$

Use Green’s theorem. The orientation is correct, and we use Green’s theorem to say the line integral is equal to $\int _{-1}^1 \int _{x^2}^1 2ydy\,dx.$

Let $C$ be the portion of $y = 2\sqrt {x}$ from $x=3$ to $x=24$ . Then $\int _C yds = \answer {156}.$

Parametrize $C$ as $\textbf {r}(t) = \left \langle t,2\sqrt {t} \right \rangle$ for $3 \leq t \leq 24$ . Then $\| \textbf {r}'(t) \| = \sqrt {1+\frac {1}{t}}$ and $f(\textbf {r}(t)) = 2\sqrt {t}$ (where $f(x,y)=y$ is the integrand). Now use the scalar line integral formula.

Let $S$ be the portion of $z = x^2+y^2$ which lies below the plane $z=1$ , oriented with outward normals. If $\textbf {F} = \left \langle x,y,1-2z \right \rangle$ , then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {\pi }.$

Let $\textbf {F} = \left \langle 2xy,x^2+3y^2 \right \rangle$ . The vector field $\textbf {F}$ isis not conservative.

If $\textbf {F} = \nabla f$ an $f(0,0) = 1$ , then $f(1,1) = \answer {3}$ .

Let $S$ be the surface given by $z = 2-\sqrt {x^2+y^2}$ with $z \geq 0$ . Then $\iint _S x^2dS = \answer {4\pi \sqrt {2}}.$

Parametrize $S$ as $\textbf {r}(r,\theta ) = \left \langle r\cos (\theta ),r\sin (\theta ),2-r \right \rangle ,$ for $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$ . Then $\textbf {r}_r = \left \langle \cos (\theta ),\sin (\theta ),-1 \right \rangle , \quad \textbf {r}_{\theta } = \left \langle -r\sin (\theta ),r\cos (\theta ),0 \right \rangle ,$ and $\| \textbf {r}_r \times \textbf {r}_{\theta } \| = \sqrt {2}\,r$ . The integral becomes $\int _0^{2\pi } \int _0^{2} r^3\cos ^2(\theta )\sqrt {2} dr\,d\theta = 4\pi \sqrt {2}.$

The parametrization $\textbf {r}(x,\theta ) = \left \langle x,x^3\cos (\theta ),x^3\sin (\theta ) \right \rangle$ for $1 \leq x \leq 2$ , $0\leq \theta \leq 2\pi$ gives the surface gotten by revolving $z = f(x)$ for $1 \leq x \leq 2$ around an axis. The function $f(x) = \answer {x^3}$ and the axis is the $x$ -axis $y$ -axis $z$ -axis .

Let $C$ be the boundary of the filled-in quadrilateral with vertices $(1,1)$ , $(1,2)$ , $(2,3)$ , and $(2,1)$ , oriented clockwise. If $\textbf {F} = \left \langle e^{x^4}+y^2,xy+\sin (\ln (y)) \right \rangle ,$ then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {8/3}$ .

Green’s theorem with wrong orientation. Notice any domain issues for $\textbf {F}$ are outside the quadrilateral, so the theorem applies. If $D$ is the filled-in quadrilateral, Green’s theorem says $\int _C \textbf {F} \cdot d\textbf {r} = -\iint _D -ydA = \iint _D ydA.$ Set up the double integral as $\int _1^2 \int _1^{x+1} ydy\,dx = 8/3.$

If $C$ is the curve given by $\textbf {r}(t) = \left \langle \cos ^3(t),\sin ^3(t),t \right \rangle$ for $0 \leq t \leq 3\pi /2$ , then $\int _C \sin (z)dx+\cos (z)dy+(xy)^{1/3}dz = \answer {1/2}.$

Do this integral directly. You should get $\int _0^{3\pi /2} \cos (t)\sin (t)dt.$

If $S$ is the portion of the plane $z = 3y$ lying over the rectangle $[0,1] \times [0,1]$ in the $xy$ -plane, oriented with upward facing normals, then $\iint _S \left \langle 0,0,z \right \rangle \cdot d\textbf {S} = \answer {3/2}.$

Let $\textbf {F} = \left \langle 1+\sin (x),x^2+e^{y^4} \right \rangle$ , and let $C$ be the curve $y =\sqrt {x}$ from $(0,0)$ to $(1,1)$ followed by the line segment from $(1,1)$ to $(1,0)$ . Then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {6/5 - \cos (1)}.$

This is a Green’s theorem + closing problem. Let $C'$ be the line segment from $(0,0)$ to $(1,0)$ . Then $-\int _C \textbf {F} \cdot d\textbf {r} + \int _{C'} \textbf {F} \cdot d\textbf {r} = \iint _D 2x dA,$ where $D$ is the region given by $0 \leq y \leq \sqrt {x}$ , $0 \leq x \leq 1$ . The double integral is $4/5$ , and the integral over $C'$ is $2 - \cos (1)$ .

Let $S$ be the portion of $z = 1-x^2-y^2$ with $z \geq 0$ , $y \geq 0$ , $x \geq 0$ , and $y \leq x/\sqrt {3}$ . Then $\iint _S \sqrt {1+4x^2+4y^2}dS = \answer {\pi /4}.$

Let $R$ be the region in the $xy$ -plane given by $x^2+(y-1)^2 \leq 1$ and $y \geq \sqrt {3}|x|$ . Then $\iint _R (2x+1)dA = \answer {\frac {1}{6}(3\sqrt {3}+2\pi )}.$

The polar set up is $\int _{\pi /3}^{2\pi /3} \int _0^{2\sin (\theta )} (2r\cos (\theta )+1)rdr\,d\theta .$

Let $S$ be the portion of $z = x^2+y^2$ lying over the triangle in the $xy$ -plane bounded by $y=0$ , $y=x$ and $x=1$ , oriented with downward pointing normals. Let $\textbf {F} = \left \langle x,z,0 \right \rangle$ . Then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {4/5}.$

Parametrize $S$ as $\textbf {r}(x,y) = \left \langle x,y,x^2+y^2 \right \rangle$ , where the domain is the triangle $T$ given in the problem. Compute $\textbf {r}_x \times \textbf {r}_y = \left \langle 2x,2y,-1 \right \rangle ,$ which is the right normal. Use the formula to compute a flux integral from here, and the setup should be $\int _0^1 \int _0^x [2x^2+2y(x^2+y^2)]dy\,dx.$

An equation of the tangent plane to the surface given by $\textbf {r}(u,v) = \left \langle u,ue^v,v \right \rangle$ at the point $(x,y,z) = (2,2e^{-1},-1)$ is $\answer {e^{-1}}(x-2) - (y-2e^{-1})+\answer {2e^{-1}}(z+1) = 0.$

If $\textbf {F} = \left \langle \frac {-y}{x^2+y^2},\frac {x}{x^2+y^2} \right \rangle$ , and $C$ is the circle $x^2+y^2 =4$ , oriented counterclockwise, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {2\pi }$ .

Notice since $\textbf {F}$ is not defined at the origin, this eliminates any hope of $\textbf {F}$ being conservative with our definition as well as any hope of using Green’s theorem. Compute the integral directly.

If $\textbf {F} = \left \langle \sin ^3(x^3)+y,x+y+1 \right \rangle$ and $C$ is the portion of $x= 1-y^2$ from $(0,-1)$ to $(0,1)$ , then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {2}$ .

Notice $\textbf {F}$ is conservative, so we can change the path to be whatever we want. Compute the integral over the line segment from $(0,-1)$ to $(0,1)$ instead, and get $2$ .