Check these two Triple Integrals (cylindrical and spherical)

Apr 2016
1
0
Colorado
Sphere Work (the picture I drew is an xy trace of the sphere, if that isn't clear)
http://i.imgur.com/WMWUp1D.jpg?1

The sphere problem asks you to find the volume of the region bounded by the plane z=1/2 and a ball with radius 1 centered at the origin using spherical coordinates. I'm not even expected to evaluate it, just figure out the bounds correctly. I guess the major work was finding the phi value of pi/3 and the rho value of 1/2sec(phi)

Cylinder Work
http://i.imgur.com/iysXAvR.jpg?1

The cylinder problem asks to find the smaller volume bounded by the cylinder of radius 4, the plane x=sqrt(8) and the plane z=100-3y. It was a little bit more complicated at least to me, I had to calculate where x=sqrt(8) intersected with the xy trace of the cylinder, define the height in cylindrical and as I'm writing this I realize I screwed up my R bounds. In order to get the minor segment of the xy trace (the smaller segment) it should be r=sqrt(8)sec(theta)..4.

I consider this post already helpful since I've found a mistake just from writing it but I would definitely appreciate someone else looking at my work briefly and make sure I'm doing these right. They evaluate out reasonably with Wolfram, so I suspect I'm at least close.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
It's always useful to think of these problems through slicing cross sections.

In the first problem, as your lower boundary is a plane (z = 1/2), then one might think to do cross-sections parallel to that plane. Every one of them is a circle. So cylindrical polars are the best option.

Clearly as each circular cross section is completely swept out, $\displaystyle \begin{align*} 0 \leq \theta \leq 2\,\pi \end{align*}$ and as you're bounded above by a unit sphere centred at the origin, the highest point will be z = 1, thus $\displaystyle \begin{align*} \frac{1}{2} \leq z \leq 1 \end{align*}$. So the hard part will be the radii. The maximum radius will be where the circular cross-sections are at their largest, i.e. where z = 1/2, so

$\displaystyle \begin{align*} x^2 + y^2 + z^2 &= 1 \\ x^2 + y^2 + \left( \frac{1}{2} \right) ^2 &= 1 \\ x^2 + y^2 + \frac{1}{4} &= 1 \\ x^2 + y^2 &= \frac{3}{4} \\ x^2 + y^2 &= \left( \frac{\sqrt{3}}{2} \right) ^2 \end{align*}$.

So the maximum radius is $\displaystyle \begin{align*} \frac{\sqrt{3}}{2} \end{align*}$. Thus $\displaystyle \begin{align*} 0 \leq r \leq \frac{\sqrt{3}}{2} \end{align*}$.

Thus the volume is $\displaystyle \begin{align*} V = \int_0^{2\,\pi}{\int_{\frac{1}{2}}^{1}{\int_0^{ \frac{ \sqrt{3} }{ 2 } }{r\,\, \mathrm{d}r }\,\, \mathrm{d}z }\,\, \mathrm{d}\theta } \end{align*}$.
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Sphere Work (the picture I drew is an xy trace of the sphere, if that isn't clear)
http://i.imgur.com/WMWUp1D.jpg?1

The sphere problem asks you to find the volume of the region bounded by the plane z=1/2 and a ball with radius 1 centered at the origin using spherical coordinates. I'm not even expected to evaluate it, just figure out the bounds correctly. I guess the major work was finding the phi value of pi/3 and the rho value of 1/2sec(phi)

Cylinder Work
http://i.imgur.com/iysXAvR.jpg?1

The cylinder problem asks to find the smaller volume bounded by the cylinder of radius 4, the plane x=sqrt(8) and the plane z=100-3y. It was a little bit more complicated at least to me, I had to calculate where x=sqrt(8) intersected with the xy trace of the cylinder, define the height in cylindrical and as I'm writing this I realize I screwed up my R bounds. In order to get the minor segment of the xy trace (the smaller segment) it should be r=sqrt(8)sec(theta)..4.

I consider this post already helpful since I've found a mistake just from writing it but I would definitely appreciate someone else looking at my work briefly and make sure I'm doing these right. They evaluate out reasonably with Wolfram, so I suspect I'm at least close.
As for the second question, again, you must think in terms of slices.

First, you're bounded below by the x-y plane (I think), so where z = 0, and you're bounded above by the plane z = 100 - 3y, thus $\displaystyle \begin{align*} 0 \leq z \leq 100 - 3y \end{align*}$.

Then if you do a slice parallel to the x-y plane, you will have a circular segment, bounded on the left by $\displaystyle \begin{align*} x = 2\,\sqrt{2} \end{align*}$ and bounded on the right by $\displaystyle \begin{align*} x = 4 \end{align*}$. As for the y's, if you realise that you will have both the top and bottom halves of the circle (at least in your region you are integrating over) then you have $\displaystyle \begin{align*} -\sqrt{16 - x^2} \leq y \leq \sqrt{16 - x^2} \end{align*}$.

Thus the volume is $\displaystyle \begin{align*} V = \int_{2\,\sqrt{2}}^{4}{\int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}}{\int_0^{100-3\,y}{\,\,\mathrm{d}z}\,\,\mathrm{d}y}\,\,\mathrm{d}x} \end{align*}$.