How to find the volume of a sphere [spherical coordinates]

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i don't know to using limit of r ?
 

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i'm sorry i mean hemi-sphere of : sqrt[x^2+y^2+(z-a)^2] = R
 
I think there are mistakes between line 3 and 4. Where did you get the subtraction from? Shouldn't it be a multiplication? Also the cosine values seem wrong. And finally, why did you stop at the easiest of all integrals at the end ##\int r^2dr## from ##r= 0## to ?
 
Please do not open more than one thread with the same topic, especially if the two are both ambiguous: with or without cone, what is ##a## needed for and what is ##r## in your other thread. Furthermore, do not delete the homework template, use it! It makes reading a lot easier and if you delete it, it can be viewed as disrespectful to those who are willing to answer.

Thread closed.
The duplicate is https://www.physicsforums.com/threads/how-to-find-hemi-sphere-on-the-cone.939376/#post-5938849
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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