Use polar coordinates to compute volumes, via the change of variables theorem

Question

So there is a question in my Anaylis book (with 3 subquestions) that I think understand, but I can not seem to understand the approach used in the solution. I've tried all subquestions, and for each of them I seem to make a mistake somewhere. When I then look at the solution they approach it differently, and I don't understand why. The question goes:

Use polar coordinates to compute

1. the volume of the region enclosed by the $x y$ -plane and the paraboloid $z=25-x^{2}-y^{2}$;
2. $\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} \frac{1}{1+x^{2}+y^{2}} d x d y$;
3. the volume of $R=\left\{(x, y, z): 0 \leq z \leq \sqrt{4-x^{2}-y^{2}},(x-1)^{2}+y^{2} \leq 1\right\}$.

To solve this I take the following steps. I sketch the volumes to get an idea of the problem. I then create my new region in polar coordinates by transforming the constraints via $x=rcos\phi$ and $y=rsin\phi$ (for $\mathbb{R^2}$, for $\mathbb{R^3}$ they're different but I won't mention them). This transforming is hard (for me) and this always seems to give me incorrect boundaries for my integrals. Then once I have the right boundaries I rewrite my integral over the new region and multiply it by the absolute Jacobian of my polar coordinate functions, which in case of polar coordinates (in $\mathbb{R^2}$) is $r$. This is because of the change of variable theorem. Then I can integrate over $\phi$ and over $r$ one by one and compute my result.

These are the provided solutions:

1: approach 1

$$ \begin{aligned} A &=\int_{0}^{2 \pi} d \varphi \int_{0}^{5} rdr\left(25-r^{2}\right) \\ &=\int_{0}^{2 \pi} 2 \varphi\left[\frac{25}{2} r^{2}-\frac{1}{4} r^{4}\right]_{0}^{5} \\ &=2 \pi\left(\frac{5^{4}-5^{4}}{4}\right)=\frac{1}{2} \pi 5^{4} \\ &=\frac{625}{2} \pi \end{aligned} $$

1. approach 2: $$ \begin{array}{l} x^{2}+y^{2}=r^{2} \\ \int_{0}^{25} d z\left(\int_{x^{2}+y^{2} \leq 25-z}d x d y\right)= \\ \int_{0}^{25} d z(\pi \cdot(25-z))= \\ \pi \cdot\left[25 z-\frac{1}{2} z^{2}\right]_{0}^{25}= \\ \pi \cdot\left(25^{2}-\frac{1}{2} 25^{2}\right)=\frac{1}{2}(25)^{2} \cdot \pi \\ =312\frac{1}{2} \pi \end{array} $$

\begin{aligned} & \int_{0}^{\pi / 2} d \varphi \int_{0}^{1} rdr\left(\frac{1}{1+r^{2}}\right) \\ =& \int_{0}^{\pi / 2} d \varphi\left[\ln \left(1+r^{2}\right)\right]_{0}^{1} \\ =& \frac{\pi}{2} \cdot \ln (2) \end{aligned}

3.

\begin{array}{l} \sqrt{4 x^{2}-y^{2}}=\sqrt{4-r^{2}} \\ (x-1)^{2}+y^{2} \leq 1 \Leftrightarrow \\ x^{2}-2 x+1+y^{2} \leq 1 \\ x^{2}+y^{2} \leq 2 x \\ r^{2} \leq 2+\cos \varphi \\ r \leq 2 \cos \varphi \end{array}

\begin{array}{l} \text { So Vol}(R)= \\ \int_{-\pi / 2}^{+\pi / 2} \int_{r=0}^{2 \cos \varphi} r d r(\sqrt{4-r^{2}})= \end{array}

\begin{array}{l} =\int_{-\pi / 2}^{\pi / 2} d \varphi\left[-\frac{1}{3}(4-r^2)^{3 / 2}\right]_{0}^{2 \cos \varphi} \\ =\int_{-\pi / 2}^{\pi / 2} d \varphi\left(4 / 3-\frac{8}{3}\left(1-\cos ^{2} \varphi\right)^{3 / 2}\right) \\ =\int_{-\pi / 2}^{\pi / 2} d \varphi\left(4 / 3-\frac{8}{3}|\sin \varphi|^{3}\right) \\ =2 \cdot \int_{0}^{\pi / 2} d \varphi\left(4 / 3-\frac{8}{3} \sin ^{3} \varphi\right) \\ =2 \cdot\left(\frac{4}{6} \pi-8 / 3-8 / 9\right)=\frac{4}{3} \pi-\frac{32}{9} \end{array}

I hope I scanned/typed the solutions correctly.

My question goes:

• Can anyone explain the steps taken to get to the boundaries in parts 1, 2 and 3?
• Where does the $r$ go in part 2 which states $rdr$ right before it is integrated?

I hope I have explained everything clearly. Cheers!

Answer 1

I would usually get the bounds for polar coordinates by sketching the shape of the region of integration and looking for a polar equation. In case 1 the region is a disk centered at the origin and in case 2 it's a quarter of a disk centered at the origin; so the limits for $r$ run from $0$ (at the origin) to the radius of the disk, and the limits for $\phi$ run all the way round the disk in case 1 and just one-quarter of the way around in case 2.

In case 3 you have a disk with the origin on the circumference. You might happen to recall that the polar equation of the circle bounding that disk is $r = 2\cos\phi.$ To "cover" the area of the disk, you need to integrate along the radials in all directions to the right of the origin: everything between $-\frac\pi2$ and $0$ to cover the region below the $x$ axis, and $0$ to $\frac\pi2$ to cover the region above the $x$ axis.

In part 2, the $r$ didn't "go" anywhere, or you might say it "went" to the same place as the $1/(1+r^2)$. The solution requires evaluating an integral in $r,$ $$ \int r \,dr\left(\frac1{1+r^2}\right) = \int \left(\frac r{1+r^2}\right) dr = \frac12\ln(1+r^2). $$

Note that the given "solution" omitted the factor $\frac12.$ The "solution" provided is therefore twice as large as the correct answer.