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Just wondering if I did this correct.

Find the volume of the region that lies inside the sphere [tex]x^2 + y^2 + z^2 = 2 [/tex] and outside the cylinder [tex]x^2 + y^2 = 1[/tex]

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Using cylindrical coordinates, and symmetry, I got:

I went up the z-axis, hitting z = 0 first, then exiting at [tex]z = \sqrt{2-r^2}[/tex]

So, the projection is two circles, one with r=1 and the other [tex]r=\sqrt{2}[/tex]

[tex]2 \int_{0}^{2\pi}\int_{1}^{\sqrt{2}}\int_{0}^{\sqrt{2-r^2}} r dz dr d\theta[/tex]

Which is then

[tex] 4\pi \int_{1}^{\sqrt{2}} r \sqrt{2-r^2} dr [/tex]

Let[tex] u = 2-r^2

=> du = -2rdr

=> \frac{-du}{2} = rdr [/tex]

Then I got:

[tex] -2\pi \int_{r=1}^{r=\sqrt{2}} \sqrt{u}du = \left[ -2\pi \frac{2}{3} u^\frac{3}{2} \right]_{r=1}^{r=\sqrt{2}}[/tex]

[tex]\left[ \frac{-4\pi}{3}(2-r^2)^\frac{3}{2} \right]_{1}^{\sqrt{2}} [/tex]

Which equals [tex]\frac{4\pi}{3}[/tex]

Look ok? Thanks.

Find the volume of the region that lies inside the sphere [tex]x^2 + y^2 + z^2 = 2 [/tex] and outside the cylinder [tex]x^2 + y^2 = 1[/tex]

----------

Using cylindrical coordinates, and symmetry, I got:

I went up the z-axis, hitting z = 0 first, then exiting at [tex]z = \sqrt{2-r^2}[/tex]

So, the projection is two circles, one with r=1 and the other [tex]r=\sqrt{2}[/tex]

[tex]2 \int_{0}^{2\pi}\int_{1}^{\sqrt{2}}\int_{0}^{\sqrt{2-r^2}} r dz dr d\theta[/tex]

Which is then

[tex] 4\pi \int_{1}^{\sqrt{2}} r \sqrt{2-r^2} dr [/tex]

Let[tex] u = 2-r^2

=> du = -2rdr

=> \frac{-du}{2} = rdr [/tex]

Then I got:

[tex] -2\pi \int_{r=1}^{r=\sqrt{2}} \sqrt{u}du = \left[ -2\pi \frac{2}{3} u^\frac{3}{2} \right]_{r=1}^{r=\sqrt{2}}[/tex]

[tex]\left[ \frac{-4\pi}{3}(2-r^2)^\frac{3}{2} \right]_{1}^{\sqrt{2}} [/tex]

Which equals [tex]\frac{4\pi}{3}[/tex]

Look ok? Thanks.

Last edited: