Integral

**vincisonfire** · December 17th 2008, 01:10 PM

I want to evaluate the volume inside $r=cos(\theta)$ .
Thanks.

**o_O** · December 17th 2008, 01:33 PM

$\begin{aligned}r & = \cos \theta \\ r^2 & = r\cos \theta \\ x^2 + y^2 & = x \\ (x-\tfrac{1}{2})^2 + y^2 & = \tfrac{1}{4} \end{aligned}$

Revolving it around the x-axis produces a sphere with radius $\tfrac{1}{2}$ .

**galactus** · December 17th 2008, 01:34 PM

That is just a circle of radius 1 centered at (1/2,0)

A sphere if you revolve

If you must do this with integration, you can just move it over the the origin.

Now, revolve a circle of radius 1/2.

$2{\pi}\int_{0}^{\frac{1}{2}}(\frac{1}{4}-x^{2})dx$

or leave it where it is and:

${\pi}\int_{0}^{1}\left(\frac{1}{4}-(x-\frac{1}{2})^{2}\right)dx$

**vincisonfire** · December 17th 2008, 02:20 PM

I want the volume within $\sqrt{x^2+y^2+z^2} = \frac{x}{\sqrt{x^2+y^2}}$ *
Here is a picture of the solid.

**badgerigar** · December 18th 2008, 12:04 AM

I want the volume within

*

I suggest you check that you have the right definition of $\theta$ because my book used it for the other angle. Anyway, using your notation:
$\int\int\int dV=\int_0^{\pi}\int_0^{2\pi}\int_0^{cos(\theta)}r^ 2\sin(\phi)dr d\theta d\phi$

**vincisonfire** · December 18th 2008, 06:09 AM

Physicists and mathematicians use the opposite definition right.
If you use the physicists' you get a sphere.
If you the mathematicians' you get this weird figure I've showed above.

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