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**Summary:**Find the volume V of the solid inside both ## x^2 + y^2 + z^2 =4## and ## x^2 +y^2 =1##

My attempt to answer this question: given ## x^2 + y^2 +z^2 =4; x^2 + y^2 =1 \therefore z^2 =3 \Rightarrow z=\sqrt{3}##

## \displaystyle\iiint\limits_R 1dV = \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi}\displaystyle\int_0^2 1 \rho^2 \sin{(\phi)} d\rho d\phi d\theta =\frac{32\pi}{3}##

Now using ##z=\sqrt{3}## we get ## \displaystyle\int_0^{2\pi}\displaystyle\int_0^\pi \displaystyle\int_0^{\sqrt{3}} 1 \rho^2\sin{\phi}d\rho d\phi d\theta = 4\sqrt{3}\pi##

Hence the area of the solid is ##\frac{32\pi}{3} -4\sqrt{3}\pi =\frac{4\pi}{3}*(8-3^{3/2})##

Actually, I couldn't draw out the graph of this solid.

I think the problem is solved now.

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