Finding the Volume of Two Solids: A Cylindrical Approach

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Hello I need help for this problem, it has been 4 hours trying to do it

Homework Statement



Find the volume of the region of space above the xy-plane, inside the cone z=7−\sqrt{x^{2}+y^{2}} and inside the cylinder x^{2}+y^{2}=4x.

Homework Equations





The Attempt at a Solution



I tried to switch to cylindrical coordinates and I got

0\leq\theta\leq2\pi
0\leqr\leq4cos(\theta)
0\leqz\leq7-r

so,
V=\int^{2\pi}_{0}\int^{4cos(\theta)}_{0}\int^{7-r}_{0}rdzdrd\theta
which doesn't work...

thanks in advance
 
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I think you'll find that θ goes from -π/2 to π/2 .
 
thank you
 
SammyS said:
I think you'll find that θ goes from -π/2 to π/2 .

How did you find these bounds?
 
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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