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

Thanks.

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- Dec 17th 2008, 01:10 PMvincisonfireIntegral
I want to evaluate the volume inside $\displaystyle r=cos(\theta) $.

Thanks. - Dec 17th 2008, 01:33 PMo_O
$\displaystyle \begin{aligned}r & = \cos \theta \\ r^2 & = r\cos \theta \\ x^2 + y^2 & = x \\ (x-\tfrac{1}{2})^2 + y^2 & = \tfrac{1}{4} \end{aligned}$

Revolving it around the x-axis produces a sphere with radius $\displaystyle \tfrac{1}{2}$ . - Dec 17th 2008, 01:34 PMgalactus
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.

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

or leave it where it is and:

$\displaystyle {\pi}\int_{0}^{1}\left(\frac{1}{4}-(x-\frac{1}{2})^{2}\right)dx$ - Dec 17th 2008, 02:20 PMvincisonfire
I want the volume within $\displaystyle \sqrt{x^2+y^2+z^2} = \frac{x}{\sqrt{x^2+y^2}} $*

Here is a picture of the solid. - Dec 18th 2008, 12:04 AMbadgerigarQuote:

$\displaystyle \int\int\int dV=\int_0^{\pi}\int_0^{2\pi}\int_0^{cos(\theta)}r^ 2\sin(\phi)dr d\theta d\phi$ - Dec 18th 2008, 06:09 AMvincisonfire
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.