Name: Calculus III
Exam 3
6 December 2000



Engage! - Captain Jean-Luc Picard


Instructions: Show all work to receive full or partial credit. You may use a scientific calculator/graphing calculator. All University rules and guidelines for student conduct are applicable.

1.

[20 pts] Calculate

$$\iint_S (x^2+y) dA$$

where S is the region between the line y=4x and the curve y=x².

The point of intersection of the two curves is easily found to be (0,0) and (4,16). Then, one can calculate the integral directly:

$$\begin{eqnarray*}\int_0^4 \int_{x^2}^{4x} (x^2+y) dy dx & = & \int_0^4 \left(x... ...3 + 8 x^2) dx \\ & = & \frac{1792}{15} \\ & \approx & 119.5 \end{eqnarray*}$$

2.

[20 pts] Find the surface area of the plane

x + 2y + 3z = 10

over the disk

$$x^2 + y^2 \leq 1.$$

The first step is to solve for z=f(x,y).

$$z = \frac{10}{3} - \frac{x}{3} - \frac{2y}{3}$$

Thus, we need to compute

$$S = \iint_R \sqrt{1+\frac{1}{9} + \frac{4}{9}} dA$$

where R is the unit disk. Since the integrand is a constant, one can quickly find S by noting that the area of the unit disk is $\pi$, and so

$$S=\frac{\sqrt{14} \pi}{3}.$$

However, if you do not see this, you can integrate $\frac{\sqrt{14}}{3}$ over the unit disk in polar coordinates to arrive at the solution.

3.

[20 pts] Find the mass of the object sketched below if $\rho = 3-z$.

$\parbox{1.5in}{\epsfxsize=1.5in \epsfbox{pyramid.eps}}$

The most direct way to solve this problem involves integrating in z first from 0 to 2, and then having the inner double integral correspond to the square shown.

$\parbox{1.5in}{\epsfxsize=1.5in \epsfbox{pyramid_soln.eps}}$

If you do the algebra, you find that the coordinate of the far corner is (1-z/2,1-z/2) so that the integral is

$$\begin{eqnarray*}m & = & \int_0^2 \int_0^{1-\frac{z}{2}} \int_0^{1-\frac{z}{2}} ... ...frac{7z^2}{4} - \frac{z^3}{4}\right) dz \\ & = & \frac{5}{3}. \end{eqnarray*}$$

4.

[20 pts] Consider a spherical mass of radius a with density

$$\sigma = a^3-(x^2 + y^2)z.$$

These problems were all straightforward if you knew the different coordinate systems. It's just a matter of changing variables and setting the appropriate limits on the integration.

(a) Set up but do not evaluate the first moment M_xy in spherical coordinates.

In spherical coordinates, the limits of integration are trivial but the translation is a bit challenging.

$$\begin{eqnarray*}M_{xy} & = & \int_0^a \int_0^\pi \int_0^{2\pi} \sigma z \rho^2 ... ...n^2\phi\cos\phi] \rho^3\cos\phi\sin\phi d\theta d\phi d\rho. \end{eqnarray*}$$

(b) Set up but do not evaluate the first moment M_xy in cylindrical coordinates.

For this one, this integrand is easier, but the limits of integration are tougher. Remember that the equation for a sphere is

x² + y² + z² = a²

or in cylindrical coordinates

r² + z² = a².

Thus,

$$M_{xy} = \int_0^{2\pi} \int_0^a \int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} (a^3-r^2z)zrdzdrd\theta.$$

(c) Set up but do not evaluate the first moment M_xy in rectangular coordinates.

This one is a natural progression from the first two.

$$M_{xy} = \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}... ...t{a^2-x^-y^2}}^{\sqrt{a^2-x^2-y^2}} (a^3-(x^2+y^2)z)zdzdydx.$$

5.

[20 pts] If

$$\v F(x,y) = \langle x,\frac{1}{y} \rangle,$$

find $\int_C \v F \cdot d\v r$ where C is the curve below

$\parbox{3in}{\epsfxsize=3in \epsfbox{path.eps}}$

A quick check reveals that this vector field has a potential. Knowing this, you can follow two approaches. One is to find the potential and use it to compute the path integral. The second approach is to use the fact that a conservative field is path independent. Thus, you do not have to follow the path shown. You can integrate from (1,1) to (4,1) in a straight line, for instance. Integrating the vector field, one find that

$$p(x,y) = \frac{1}{2} x^2 + \ln \vert y\vert.$$

Thus,

$$\int_C \v F \cdot d\v r = p(4,1)-p(1,1) = \frac{15}{2}.$$