[Math] Volume of cylinder inside of sphere

cylindrical coordinatesintegrationmultiple integralspherical coordinates

I've bee given 2 equations in spherical form. One of a sphere $(s_1)$ and the other of a cylinder $(s_2)$.

$$S_1 : \rho = 4\cos \phi \text{ and } S_2 : \rho \sin \phi = 1$$

I need to find the volume inside the sphere and inside the cylinder using both spherical and cylindrical coordinates.

So far I was able to convert surfaces to Cartesian form

$$S_1 : x^2 + y^2 + (z-2)^2 = 4 \text{ and } S_2: x^2 + y^2 = 1$$

But now when I try calculating the volume, I get values that dont make sense.

Im expecting a volume of a bit less than $4\pi $ $units^3$ $(4 \cdot \pi \cdot
1^2)$, but I cant seem to come near that. Im getting values around $5$ and $6$ (depending on if I calculated it in cylindrical or cartesian)

Please help. Thanks

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Best Answer

The equation of the sphere is $r^2+(z-2)^2=4$, so $z=2 \pm \sqrt{4-r^2}$. The solid of interest is bounded below and above by the sphere and by the cylinder $r=1$ on the side.

Then,

$$V=\int_{0}^{2\pi} \int_{0}^{1} \int_{2-\sqrt{4-r^2}}^{2+\sqrt{4-r^2}} r dz dr d\theta$$

$$=\int_{0}^{2\pi} \int_{0}^{1} 2r\sqrt{4-r^2} dr d \theta$$

$$=-2\pi \int_{4}^{3} u^{\frac{1}{2}} du$$

$$=\frac{4}{3}\pi (8-3\sqrt{3})$$

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