Second Midterm Midpoint Assessment

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General Concepts

If $f(x,y)$ is a continuous function on a region $D$ in the $xy$ -plane, then $\displaystyle \iint _D f(x,y)dA$ represents the of the region below the graph of $z = f(x,y)$ lying over the region $D$ .

If $f(x,y)$ is a continuous function and $C$ is a curve in the $xy$ -plane, then $\displaystyle \int _C f(x,y)ds$ is called a and represents the of the region under $z = f(x,y)$ lying over the curve $C$ .

Which of the following functions is an example of a vector field?

Give the definition of a conservative vector field.

What does a vector line integral $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ represent geometrically?

If $R$ is a region in the $xy$ -plane, then what does $\displaystyle \iint _R 1dA$ represent?

If $C$ is a curve, then $\displaystyle \int _C 1ds$ gives what?

True/False Questions

Answer each of the following as true or false, and explain your answe. Unless otherwise specified, all functions and all components of vector fields are defined and continuous everywhere.

If $C$ is an oriented curve and $\textbf {F}$ is a vector field that, at every point on $C$ , makes an acute angle with the tangent vectors of $C$ , then $\displaystyle \int _C \textbf {F} \cdot dr > 0$ .

If $f(x,y)$ is a continuous function and $C$ is a closed curve in the $xy$ -plane, then $\displaystyle \int _C f(x,y)ds = 0$ .

The vector field $\textbf {F}(x,y) = \left \langle \sin (x^2),\sin (y^2) \right \rangle$ is conservative.

If $\textbf {F} =\left \langle F_1,F_2 \right \rangle$ is a conservative vector field, then $\frac {\partial F_1}{\partial y} = \frac {\partial F_2}{\partial x}$ .

If $C$ is the unit circle, oriented counterclockwise, and $\textbf {F}$ is a vector field with $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = 0$ , then $\textbf {F}$ is a conservative vector field.

If $C$ is the unit circle, oriented counterclockwise, and $\textbf {F}$ is a vector field with $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = 0$ , then $\textbf {F}$ is perpendicular to the curve $C$ at every point on the circle.

If $\textbf {F}$ is a conservative vector field, then $\textbf {F}$ has infinitely many potential functions $f$ .

If $f(x,y)$ has continuous second partial derivatives on $\mathbb {R}^2$ , and $C_1$ and $C_2$ are two paths from $(0,0)$ to $(2,1)$ , then $\displaystyle \int _{C_1} \nabla f \cdot d\textbf {r} = \int _{C_2} \nabla f \cdot d\textbf {r}$ .

If $C$ is an oriented curve and $C'$ is the curve $C$ with the opposite orientation, and $\textbf {F}$ is a vector field, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}= \int _{C'} \textbf {F} \cdot d\textbf {r}$ .

If $C$ is an oriented curve in the $xy$ -plane and $C'$ is the curve $C$ with the opposite orientation, and $f(x,y)$ is an everywhere continuous function, then $\displaystyle \int _C f(x,y)ds = \int _{C'} f(x,y)ds$ .

If $\textbf {F} = \left \langle -y,0 \right \rangle$ , and $C$ is a closed curve (oriented counterclockwise) which encloses a region $R$ in the $xy$ -plane, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \text {area}(R)$ .

If $C$ is a curve from point $P$ to point $Q$ in the $xy$ -plane, and $f(x,y)$ is a continuous function, then $\displaystyle \int _C f(x,y)ds = f(Q) - f(P)$ .

Free Response

If $f(x,y) = (x+2y)^2$ and $R$ is the rectangle $[0,1] \times [-1,1]$ in the $xy$ -plane, then $\displaystyle \iint _R f(x,y)dA = \answer {10/3}$

The volume of the solid bounded by $z = x^2+y^2$ and the planes $z = 0$ , $y = 0$ , $x=0$ , and $x+y=1$ is: $\answer {1/6}$

If $R$ is the region given by $y \leq \sqrt {1-x^2}$ , $y \geq x$ , and $y \geq -x$ , and $f(x,y) = y$ , then $\displaystyle \iint _R f(x,y)dA = \answer {\frac {\sqrt {2}}{3}}$ .

The volume of the solid bounded by $z = 8-x^2-y^2$ and $z = x^2+y^2$ is: $\answer {16 \pi }$ .

The integral $\int _0^4 \int _{\sqrt {y}}^2 \sin (x^3)dxdy = \answer {-\frac {1}{3}\cos (8)+\frac {1}{3}}$

If $C$ is the curve given by $\textbf {r}(t) = \left \langle 2t,3t,4t \right \rangle$ for $0 \leq t \leq 1$ , and $f(x,y,z) = z^2$ , then $\displaystyle \int _C f(x,y,z)ds = \answer {16\sqrt {29}/3}$ .

The integral $\displaystyle \int _{-1/\sqrt {2}}^0 \int _{-x}^{\sqrt {1-x^2}} (x-3)dydx$ in polar coordinates is: $\int _{\answer {\pi /2}}^{\answer {3\pi /4}} \int _{\answer {0}}^{\answer {1}} \answer {(r\cos (\theta )-3)r} dr\,d\theta ,$ and the integral is equal to: $\answer {\frac {1}{24}(-8+4\sqrt {2}-9\pi )}$ .

If $\textbf {F}(x,y) = \left \langle x+y,y-x \right \rangle$ and $C$ is the line segment from $(0,0)$ to $(1,2)$ , then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {5/2}$ .

If $\textbf {F} = \left \langle y,x \right \rangle$ and $C$ is the line segment from $(1,-1)$ to $(2,-3)$ , then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {-5}$ .

If $\textbf {F} = \left \langle x+y,y-x \right \rangle$ and $C$ is the circle $x^2+y^2=1$ oriented clockwise, then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {2\pi }$ .

Let $R$ be the region in the first quadrant inside both the circle $x^2+y^2=1$ and the circle $(x-1)^2+y^2=1$ . Then $\displaystyle \iint _R ydA = \answer {5/24}$ .

If $\textbf {F} = \left \langle x^2,3x,xz \right \rangle$ and $C$ is the line segment from $(1,2,3)$ to $(1,0,4)$ , then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {-5/2}$

Let $C$ be the cure going from $(0,0)$ to $(2,1)$ to $(5,1)$ to $(5,0)$ back to $(0,0)$ along straight line segments. Then $\int _C (x^3+e^{x+y}+e^{x^2})dx + (xy+e^{y^2}+e^{x+y})dy = \answer {-\frac {11}{6}}.$

If $C$ is the curve given by $\textbf {r}(t) = \left \langle 1+\sin ^3(t),1-\cos (t),t/\pi \right \rangle$ , with $0 \leq t \leq \pi /2$ , then $\int _C (yz+z\sin (xz))dx+(xz+y^3)dy+(xy+x\sin (xz))dz = \answer {\frac {9}{4}-\cos (1)}.$

Let $\textbf {F} = \displaystyle \left \langle x^2,e^{\cos (y)+y} \right \rangle$ and let $C$ be the arc of the parabola $y = 3x^2$ from $(1,3)$ to $(-1,3)$ . Then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = \answer {-2/3}$ .

Let $C$ be the curve given by $\textbf {r}(t) = \left \langle \cos (t),t^2,\sin (t) \right \rangle$ for $-\pi \leq t \leq 0$ . Then $\displaystyle \int _C \sqrt {y}ds = \answer { \frac {1}{12}(1+4\pi ^2)^{3/2} - \frac {1}{12}}$ .

Let $C$ be the portion of $x^2+y^2=1$ where $x \geq 0$ , oriented from $(0,1)$ to $(0,-1)$ , and let $\textbf {F} = \left \langle y^2+y\sin (xy)-\sin (x^4),y^2+x\sin (xy) \right \rangle$ . Then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {-\frac {2}{3}}$

If $C$ is the top half of $x^2+y^2=1$ oriented counterclockwise, then $\int _C \left (\cos (x)\cos (y)-\frac {y^2}{2}\right )dx + (2xy+\sin (y^6)-\sin (x)\sin (y))dy = \answer {2-2\sin (1)}.$

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