Finding Volume Using Triple Integrals: A Brief Guide

synergix
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Homework Statement



i have to find the volume between the function z=4-x^2-y^2 and the x/y plane

The Attempt at a Solution



I think I should be fine with the limits of integration but am not 100% confident what I am integrating.

is it 4-x^2-y^2-z??

or 4-x^2-y^2?
 
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So I'm just going to integrate dv I think this might be correct. my bounds would be x=0->sqrt(y^2-4) then y=0->sqrt(z-4) then z=0->4. And then I would integrate in this order. Is this correct?
 
Why not convert to polar coordinates? Then z=4-r^2 right? I assume you mean positive z, the paraboloid above the x-y plane. It's symmetrical so you could just integrate in the first quadrant and multiply by four:

V=4\int_0^{\theta_0}\int_0^{r_m} (4-r^2)rdrd\theta

Can you understand how I did that and come up with the values for \theta_0 and r_m?
 
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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