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18.02 Exam 3 -- April 9, 1999

1.

Compute the volume of the region between the $xy$ plane and
the surface $z= -x^2-y^2 +4$.

Solution: The volume is given by the double integral over the region, $R=\{x^2+y^2\le 4\}$

$$\begin{multline*} \iint_R (4-x^2-y^2)dA=\int_0^{2\pi}\int_0^{2}(4-r^2)r\, d\theta\,dr\\ =2\pi\left[2r^2-\frac{r^4}4\right]_0^2=8\pi. \end{multline*}$$

2.

Evaluate the integral

$$\int^1_0 \int^1_x \cos (y^2) \, dy \, dx \, .$$

Solution: Change the order of integration and the integral becomes

$$\int^1_0 \int^y_0 \cos (y^2) \, dx \, dy= \int_0^1y\cos(y^2)dy=\left[\frac12\sin(y^2)\right]^1_0=\frac12\sin1.$$

3.

Using polar coordinates evaluate the integral

$$\int\int_R \exp (x^2 + y^2) \, dA$$

where $R$ is the disc of radius $1$ and center the origin.

Solution: In polar coordinates the integral is

$$\int_0^{2\pi}\int_0^1 \exp (r^2) \, rdr\,d\theta = 2\pi\left[\frac12\exp(r^2)\right]_0^1=\pi(e-1).$$

4.

(a)

For what values of the constant $a$ is

$${\bf F} = (4 x^3 + 2xy + ay^2){\bf i} + (x^2 + 2y){\bf j}$$

a conservative vector field?

(b)

For $a=0$ compute

$$\int_C {\bf F} \cdot \, d {\bf r}$$

where $C$ is the semicircule with radius $1$, center $(0,0)$ from $(1,0)$ to $(-1,0)$ in the upper half-plane.

Solution:

(a)

$$\frac{\partial}{\partial y}(4 x^3 + 2xy + ay^2)=2x+2ay= \frac{\partial}{\partial x}(x^2 + 2y)=2x$$

exactly when $a=0.$

(b)

Since the vector field is conservative the integral is the same for any contour with these two endpoints, for instance the segment $[-1,1]$ of the x-axis. The integral along the x-axis is $\int_{-1}^14x^3dx=0.$

5.

(a)

For the vector field ${\bf F} =y(x^2+y^2)^2{\bf i}- x(x^2+y^2)^2{\bf j}$ compute $\operatorname{div}({\bf F})$

(b)

For this vector field and the curve $C$ which is the circle of radius $2$ with center the origin and positive orientation, use Green's theorem to compute the flux integral

$$\oint_C {\bf F} \cdot \, {\bf {\hat n}} \, ds \, .$$

Solution:

(a)

$$\begin{multline*} \operatorname{div}({\bf F})=\frac{\partial}{\partial x}[y (x^2... ...al}{\partial y}[x(x^2 + y^2)^2]\\ =4xy(x^2+y^2)-4xy(x^2+y^2)=0. \end{multline*}$$

(b)

0

6.

Sketch the parameterized curve

$$x=\cos t \, , \,\ y=\sin^3t \, , 0 \leq t \leq 2 \pi \, .$$

(a)

Why is it closed?

(b)

Why is it simple?

(c)

Use Green's theorem to express the area inside the curve as a single integral. Do not evaluate it.

OR simply:

Use Green's theorem to express the area inside the simple closed curve

$$x=\cos t \, , \,\ y=\sin^3t \, , 0 \leq t \leq 2 \pi,$$

as a single integral. Do not evaluate it.

Solution: [There is of course a small conceptual problem here that the curve is not smooth, but I don't suppose that anyone will notice.]

(a)

$x(2\pi)=x(0),$ $y(2\pi)=y(0).$

(b)

$\sin t$ takes each value between $-1$ and $1$ twice for $t$ in the interval $[0,2\pi]$ but at these points $\sin t,$ and hence $\sin^3t$ has opposite sign.

(c)

$$\oint_C -y \, dx = \int^{2 \pi}_0 \sin^4t\,dt.$$

7.

Let $R$ be the region in the first quadrant $(x \geq 0, y\ge0)$ where $3x+y \leq 1.$ Using the change of variable formula for double integrals rewrite the integral

$$\int_R \int (3x+y)^4 \exp(x+y) \, dA$$

as an iterated integral in terms of the new variables $u=3x+y$ and $v=x+y.$ Give limits of integration but do not evaluate.

Solution: For $u=3x+y$ and $v=x+y$ the Jacobian is

$$\frac{\partial (u,v)}{\partial (x,y)}= \left\vert\begin{matrix}3&1\\ 1&1 \end{matrix}\right\vert=2,$$

so ${\partial (x,y)}/{\partial (u,v)}=1/2.$ On $x=0,$ $v=u$ and on $y=0,$ $v=u/3$ so the integral becomes

$$\frac12\int^1_0 \int^{u}_{u/3} u^4 \exp(v) \, dv \, du.$$