Polar integral

[aronclark1017]

TL;DR: I'm still debating on why exactly this is not compiling on -pi/2, pi/2. It has been suggested in other post that the square root of sin^2(t) on -pi/2 is to blame. However it looks to me that the polar equation can only exist at -pi/2 on second iteration causing some type of mysterious error.

 

[BvU]

Hi,
may I remind you of the

Homework Help Guidelines for Students and Helpers​

and ask for a few clarifications ?

I spent some time finding your problem in my son's copy of
Stewart 5th ed Ch 15.8 exercise 14:

(a) Find the volume of the solid that the cylinder $r=a\cos\theta$ cuts out of the sphere of radius $a$ centered at the origin.
(b) Illustrate the solid of part (a) by graphing the sphere and the cylinder on the same screen.

Potential helpers have to guess what you are doing and with what tool you run into your 'mysterious error'. Please help us help you !

$\$

 

[Mark44]

It has been suggested in other post that the square root of sin^2(t) on -pi/2 is to blame.

How so? $\sin^2(t) \ge 0$ for all real numbers t. In particular, $\sin^2(-\pi/2) = 1$.

 

[Mark44]

In a more mathematical form that is easier to read, your first integral is:
$$2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos(t)}\int_0^{a^2 - r^2}r~dzdrdt$$
Why do you have the constant 2 in front of the leftmost integral? The circle at the base of the cylinder is completely defined as t ranges from $-\pi/2$ to $\pi/2$.

In addition, your work is extremely difficult to follow. It would be very helpful if you wrote your integrals and other mathematical stuff using LaTeX, similar to what I did above.
A few minutes studying our LaTeX Guide (the link is at the lower left) and applying what's there would probably cause more members to look at your posts.

Note that what I wrote is the following, in raw form:
2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos(t)}\int_0^{a^2 - r^2}r~dzdrdt

 

[pasmith]

In a more mathematical form that is easier to read, your first integral is:
$$2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos(t)}\int_0^{a^2 - r^2}r~dzdrdt$$
Why do you have the constant 2 in front of the leftmost integral? The circle at the base of the cylinder is completely defined as t ranges from $-\pi/2$ to $\pi/2$.

Based on the sketch, it seems that $\theta$ is the angle of cylindrical polar coordinates, not spherical polar coordinates. I agree that the base of the cylinder is completely described by $-\pi/2 \leq \theta \leq \pi/2$. The factor of 2 comes not from rotational symmetry about the origin (there isn't any) but from reflection in the plane $z = 0$: we want the volume of intersection of the cylinder and the sphere, and that extends into negative $z$.

The problem comes in line 6 of the OP's work which I interpret as $$\int_{-\pi/2}^{\pi/2} \left[ \tfrac23 u^{3/2}\right]_{a^2 \sin^2 \theta}^{a^2}\,d\theta = \frac{2a^3}3 \int_{-\pi/2}^{\pi/2} 1 - \sqrt[3/2]{\sin^2\theta}\,d\theta.$$ where we end up with an integrand involving $\sqrt[3/2]{\sin^2\theta}$. This is not $\sin^3 \theta$, as the OP has assumed, but $|\sin^3 \theta|$ since $\sin\theta$ is negative for $-\pi/2 \leq \theta \leq 0$. I suspect this is where the OP has gone astray, although I'm not sure since their working is hard to follow. However, by evenness of $\sin^2 \theta$ we should now be at the point $$
\frac{4a^3}3 \int_0^{\pi/2} 1 - \sin^3 \theta\,d\theta.$$

In addition, your work is extremely difficult to follow. It would be very helpful if you wrote your integrals and other mathematical stuff using LaTeX, similar to what I did above.

I concur.

 

[BvU]

Can't help it ...

looks as if you want to integrate $$\int\limits_{-\pi/2}^{\pi/2}\ \int\limits_0^{a\cos\theta} \int\limits_0^\sqrt{a^2-r^2} \ r\ dz\,dr\,d\theta $$ Well, copy/paste the $\LaTeX$ code to Wolframalpha (with a=1 out of pure ignorance) and sure enough:

which is exactly one half of ##\frac 4 3 a^ 3\pi\left(\frac 1 2-\frac 2 {3\pi}\right ) ## or -- when typeset -- $\frac 4 3 a^ 3\pi\left(\frac 1 2-\frac 2 {3\pi}\right )$

I'm impressed Wolfie expertly ignores the spacing I inserted for good measure...

Still curious about the tool you use ...

(where the graphics leave much to be desired ! Anyone who can suggest a freeware solids manipulation tool to visualize the intersection of a sphere and a cylinder ?)

$\$

 

[renormalize]

I'm impressed Wolfie expertly ignores the spacing I inserted for good measure...

Just want to point out that the full version of Mathematica can evaluate the triple integral symbolically for arbitrary positive $a\,$:

 

[pasmith]

Any CAS which knows that for $a \in \mathbb{R}$, $(a^2)^{1/2} = |a|$ should be able to get it right.

 

[Mark44]

Please be advised that the OP here has "left the room."