Polar Coordinates, intersection of a cylinder with a spher

[DarthRoni]

Homework Statement

Find the Volume of the solid that the cylinder $r = acos\theta$ cuts out of the sphere of radius a centered at the origin.

Homework Equations

The Attempt at a Solution

I have defined the polar region as follows,
$$D = \{ (r,\theta) | -\pi/2 ≤ \theta ≤ \pi/2 , 0 ≤ r ≤acos\theta \} $$

I will take f(x,y) as follows,
$$ f(x,y) = \sqrt{a^2 - x^2 - y^2} $$

Because of symmetry, I should be able to take the removed volume as double the double integral of f(x,y) over the region D.

$$ V = 2 *\int_{-\pi/2}^{\pi/2} \int_{0}^{acos\theta} r\sqrt{a^2 - r^2} \mathrm{d} r \mathrm{d} \theta $$

After solving this double integral, I keep ariving at ${2\pi a^3}/3$ Which I don't think makes sense because that would be half the volume of the sphere. What am I doing wrong?

 

[Ray Vickson]

Homework Statement

Find the Volume of the solid that the cylinder $r = acos\theta$ cuts out of the sphere of radius a centered at the origin.

Homework Equations

The Attempt at a Solution

I have defined the polar region as follows,
$$D = \{ (r,\theta) | -\pi/2 ≤ \theta ≤ \pi/2 , 0 ≤ r ≤acos\theta \} $$

I will take f(x,y) as follows,
$$ f(x,y) = \sqrt{a^2 - x^2 - y^2} $$

Because of symmetry, I should be able to take the removed volume as double the double integral of f(x,y) over the region D.

$$ V = 2 *\int_{-\pi/2}^{\pi/2} \int_{0}^{acos\theta} r\sqrt{a^2 - r^2} \mathrm{d} r \mathrm{d} \theta $$

After solving this double integral, I keep ariving at ${2\pi a^3}/3$ Which I don't think makes sense because that would be half the volume of the sphere. What am I doing wrong?

Show your work after the above; we cannot tell where you went wrong if you don't show us what you did---in detail, step-by-step.

 

[DarthRoni]

$V = -2/3 * \int_{0}^{2\pi} (a^2 - a^2 cos^2(\theta))^{3/2} - a^{3/2} d\theta$
$\Rightarrow V = -2/3 * \int_{-\pi/2}^{\pi/2} a^3 (1- cos^2(\theta))^{3/2} - a^3 d\theta$
$\Rightarrow V = -(2*a^3)/3 * \int_{-\pi/2}^{\pi/2} sin^3(\theta) - 1 d\theta$
$\Rightarrow V = (2*a^3*\pi)/3$

Hope that's enough, any help with be very appreciated.

 

[Ray Vickson]

$V = -2/3 * \int_{0}^{2\pi} (a^2 - a^2 cos^2(\theta))^{3/2} - a^{3/2} d\theta$
$\Rightarrow V = -2/3 * \int_{-\pi/2}^{\pi/2} a^3 (1- cos^2(\theta))^{3/2} - a^3 d\theta$
$\Rightarrow V = -(2*a^3)/3 * \int_{-\pi/2}^{\pi/2} sin^3(\theta) - 1 d\theta$
$\Rightarrow V = (2*a^3*\pi)/3$

Hope that's enough, any help with be very appreciated.

One obvious problem is the fact that $\int_{-\pi/2}^{\pi/2} \sin^3 (\theta) \, d \theta = 0$ because $\sin$ is an odd function. This, in turn, arises from the mistake of taking $(1-\cos^2)^{3/2} = \sin^3$. The fact is that you started off with positive square roots, then got a 3/2 power via integration. In fact, $\sqrt{1-\cos^2} = |\sin|$. You could have avoided all that by using symmetry to just integrate from 0 to $\pi/2$, then multiply the result by 2.

 

[DarthRoni]

Thank you very much, I see where I went wrong. After using your method, I arrived at a final answer of
$V = 4/3 a^3 *((3\pi - 4)/6)$
which is about 29 percent of the sphere's volume. Makes more sense. Thanks again for your help!