Variable substitution, help with understanding

Question

this one should be fairly easy (I hope)

I am stumped trying to figure out how this limit and variable substitution worked in detail, I am working with calc III and an intersection between two spheres, and I found this topic here on stack, Find volume between two spheres using cylindrical & spherical coordinates

Now, the specific issue I am having is interpreting the variable sub here and how it works, I have different numbers myself but If I can understand this I can solve my problem too =)

$V=2\pi \int_{0}^{\sqrt{\frac{3}{4}}} [2{\sqrt{1-r^2}}-1] rdr$

$u = 1-r^2 ; r = 0 => u = 1 ; r = \sqrt{\frac{3}{4}} => u = \frac{1}{4}$ (corrected from the post)

$V = 2\pi [-\int_1^{\frac{1}{4}} u^{\frac{1}{2}} du - \int_{0}^{\sqrt{\frac{3}{4}}} rdr]$

$V= 2\pi (\frac{2}{3}u^{\frac{3}{2}}) - (\frac{r^2}{2})$

Could anyone lead me throught the finer details here..

.. this might be easy but to me it feels convoluted and strange =) could anyone help?

My own tries are below however I am working with

$V=2\pi \int_{0}^{\sqrt{3}} [2{\sqrt{4-r^2}}-2] rdr$

http://prntscr.com/b3apr4

my answer should come out to $\frac{10\pi}{3}$ but I am not getting there, do you have insight? =)

For reference I am starting with $x^2+y^2+z^2=4$ and $x^2+y^2+(z-2)^2=4$.

Answer 1

Ok, so I found the answer =), my error was in the limits!

instead of 0-2 I needed 1-4! I was not calculating the limits properly, my bad, I found it and dealt with it.

Answer 2

It looks like they did a u subsitution and an integral spiit in the same step, causing confusion. First, take that integral in the first line and distribute the r. Then, split the integral. The integral on the right is easy, the integral on the left requires more work. For the integral on the left, substituting u=1-r^2 yields du=-2rdr. Just substitute the 1-r^2 for u and the rdr for -du/2.