Volume bounded a cone and a sphere

[dunkelheit]

Evaluate the volume of [MATH]S=\{(x,y,z)\in\mathbb{R}^3 \ \text{s.t.} \ x^2+y^2 \leq z^2, x^2+y^2+z^2 \leq z\}[/MATH].
We have that

[MATH]\text{Vol} (S) =\iiint_S \text{d}x \text{d}y \text{d}z[/MATH]​

I thought about integrating in this way

[MATH]\text{Vol} (S) =\int_S \text{d}x \text{d}y \text{d}z=\int_{z_1}^{z_2} \left(\iint_{S_z} \text{d}x \text{d}y \right) \text{d}z[/MATH]​

With [MATH]S_z=\{(x,y)\in\mathbb{R}^2 \ \text{s.t.} \ x^2+y^2 \leq z^2, x^2+y^2 \leq z-z^2\}[/MATH]The point is that I'm not sure which surface is above the other: from the graph it is obvious that it is the sphere [MATH]x^2+y^2+(z-1/2)^2 \leq 1/4[/MATH], but how do I find this with the inequalities? Furthermore, the layer varies with [MATH]z[/MATH] so, since they intersect at [MATH]z=1/2[/MATH], I suppose that I have to do something like this

[MATH]\int_{z_1}^{z_2} \left(\iint_{S_z} \text{d}x \text{d}y \right) \text{d}z=\int_0^{1/2} \pi z^2 \text{d}z+\int_{1/2}^{z^{*}} \pi(z-z^2) \text{d}z[/MATH]​

Because the layers are circles with radius [MATH]z[/MATH] and [MATH]\sqrt{z-z^2}[/MATH].
And I'm not sure on who is [MATH]z^{*}[/MATH], I think I have the limitation on [MATH]z[/MATH] by letting [MATH]x=y=0[/MATH] in [MATH]x^2+y^2+z^2 \leq z[/MATH] and so the upper bound for [MATH]z[/MATH] is [MATH]z^{*}=1[/MATH]; hence

[MATH]\int_0^{1/2} \pi z^2 \text{d}z+\int_{1/2}^{z^{*}} \pi(z-z^2) \text{d}z=\int_0^{1/2} \pi z^2 \text{d}z+\int_{1/2}^{1} \pi(z-z^2) \text{d}z=\frac{\pi}{24}+\frac{\pi}{12}=\frac{3}{24}\pi[/MATH]​

Is this correct? If it isn't, where are the mistakes? Thanks.

 

[LCKurtz]

You need to set this up in spherical coordinates. The sphere boundary is [MATH]\rho^2 = \rho\cos\phi[/MATH] or [MATH]\rho = \cos\phi[/MATH] and the cone boundary is [MATH]\phi = \frac \pi 4[/MATH]. You want the volume above the cone and under the sphere. Put the appropriate limits on
[MATH]\int\int\int \rho^2\sin\phi~d\rho d\phi d\theta[/MATH]You should get [MATH]\frac \pi 8[/MATH], I think.

 

[dunkelheit]

I actually get [MATH]\frac{\pi}{8}[/MATH], I just forgot to simplify: [MATH]\frac{3}{24} \pi=\frac{\pi}{8}[/MATH]; I was unsure about the logic, thanks for your answer because the result is the same so it's likely correct.

 

[LCKurtz]

But you should notice how much easier it is in spherical coordinates, right?

 

[dunkelheit]

Yeah, I must admit that!