## Assignment 7 Solutions

16.5 #34: Evaluate the integral

$\int_0^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-x^2-z^2}}^{\sqrt{1-x^2-z^2}}\frac{1}{(x^2+y^2+z^2)^{1/2}}\,dy\,dz\,dx.$

Solution: The bounds on this integral tell us that we’re integrating over half the sphere of radius 1 centered at the origin (specifically the hemisphere where $x>0$).  Since $\rho$ represents distance from the origin, our function becomes

$\frac{1}{(x^2+y^2+z^2)^{1/2}}=\frac{1}{\rho}.$

Thus, our integral in spherical coordinates becomes

$\int_{-\pi/2}^{\pi/2}\int_0^\pi\int_0^1 \frac{1}{\rho}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$

Now, we can use our trick of viewing $\sin\phi$ as a constant in $\rho$, and so on. We can immediately rewrite the integral as

$\int_{-\pi/2}^{\pi/2}\,d\theta\int_0^\pi\sin\phi\,d\phi\int_0^1\rho\,d\rho.$

Evaluating the integral gives us

$\pi\left(-\cos\theta\Big|_0^\pi\right)\left(\frac{1}{2}\rho^2\Big|_0^1\right)=\pi.$

16.5 #48: The density of a solid sphere at any point is proportional to the square of the distance of the point to the center of the sphere. What is the ratio of the mass of a sphere of radius 1 to a sphere of radius 2?

Solution: There is an easy and a hard way to solve this. In spherical coordinates, the mass of a sphere of radius $r$ with the given density property (density is given by $\rho^2$ where $0\le\rho\le r$) is

$\int_0^{2\pi}\int_0^\pi\int_0^r \rho^2\rho^2\sin\phi\,d\rho d\phi d\theta = \int_0^{2\pi}d\theta\int_0^\pi\sin\phi d\phi\int_0^r \rho^4 d\rho.$

What we want to do is to take the ratio of a sphere of radius 1 and a sphere of radius 2 using this formula. We get

$\frac{\displaystyle \int_0^{2\pi}d\theta\int_0^\pi\sin\phi d\phi\int_0^1 \rho^4 d\rho}{\displaystyle \int_0^{2\pi}d\theta\int_0^\pi\sin\phi d\phi\int_0^2 \rho^4 d\rho} = \frac{\displaystyle \int_0^1 \rho^4 d\rho}{\displaystyle \int_0^2 \rho^4 d\rho}.$

This is nice because now we only need to take a single antiderivative, and evaluate it on two intervals. This becomes

$\frac{\displaystyle \frac{1}{5}\rho^5\Big|_0^1}{\displaystyle \frac{1}{5}\rho^5\Big|_0^2}=\frac{\displaystyle\rho^5\Big|_0^1}{\displaystyle \rho^5\Big|_0^2}=\frac{1}{2^5}=\frac{1}{32}.$

17.1 #58: For $t>0$, a particle moves along the curve $x=a+b\sin kt$, $y=a+b\cos kt$ where $a,b,k$ are positive constants.

(a) Describe the motion in words.

Solution: The particle revolves around a circle located at $(a,a)$ having radius $b$ and period $2\pi/k$. Also, the particle starts at $(a,a+b)$ and revolves clockwise.

(b) What is the effect on the curve of the following changes? (i) Increasing $b$. (ii) Increasing $a$. (iii) Increasing $k$. (iv) Setting $a$ and $b$ equal.

Solution: Increasing $b$ will increase the radius of the circle. Increasing $a$ will move the center of the circle along the line $y=x$ farther from the origin. Increasing $k$ will make the particle travel faster around the circle. Setting $a$ and $b$ equal means that the radius of the circle is equal to the horizontal and vertical distance from the origin, meaning that the circle will rest on the $x$ and $y$-axes.

17.1 #64: Consider the line $x=5-2t$, $y=3+7t$, $z=4t$ and the plane $ax+by+cz=d$. All of the following questions have many possible answers. Find values of $a$, $b$, $c$, and $d$ such that

(a) The plane is perpendicular to the line.

Solution: The line contains the point $(5,3,0)$, and more importantly we know that

$\frac{dx}{dt}=-2,\qquad\frac{dy}{dt}=7,\qquad \frac{dz}{dt}=4.$

That is, the vector $-2\vec{i}+7\vec{j}+4\vec{k}$ must be perpendicular to every displacement vector in the plane. Alternatively, this vector is a normal vector to the plane. Since we’re not wishing to orient the plane in any specific location, we may use the origin as a point in the plane. We have that the plane can be represented as

$-2(x-0)+7(y-0)+4(z-0)=0$

and so we have $-2x+7y+4z=0$, giving $a=-2$, $b=7$, $c=4$ and $d=0$.

(b) The plane is perpendicular to the line and through the point $(5,3,0)$.

Solution: This is a bit more complicated, but we can set up the initial equation as

$-2(x-5)+7(y-3)+4(z-0)=0.$

This tells us that $-2x+10+7y-21+4z=0$, and hence $-2x+7y+4z=11$. Thus, we take our previous answer and change $d$ to $11$.

(c) The line lies in the plane.