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Use cylindrical coordinates to find the volume of the region bounded below by the plane z=1 and above by the sphere x^2+y^2+z^2=9.

User Genius
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Answer:


\displaystyle V=(28\pi)/(3)

Explanation:

The first thing we want to do is convert (x,y,z) to (r,θ,z):


x=r\cos\theta\\y=r\sin\theta\\z=1

With that out of the way, we need our bounds for each variable.

The variable
\theta is pretty easy since our region is a portion of a sphere, so with a circular cross-section, the bounds are
0\leq\theta\leq2\pi.

Now, we find the bounds for
r given the plane
z=1:


x^2+y^2+z^2=9\\(r\cos\theta)^2+(r\sin\theta)^2+1^2=9\\r^2\cos^2\theta+r^2\sin^2\theta+1=9\\r^2+1=9\\r^2=8\\r=√(8)

Therefore, the bounds for
r are
0\leq r\leq √(8) thanks to the bounds from
\theta earlier.

Lastly, with
z, we want to write it in terms of
r and get the bounds:


x^2+y^2+z^2=9\\r^2+z^2=9\\z^2=9-r^2\\z=√(9-r^2)

The radical is positive since the region is above the plane.

Now, we can form our triple integral and evaluate it:


\displaystyle \int^(2\pi)_0\int_0^(√(8))\int_1^(√(9-r^2))r\,dz\,dr\,d\theta\\\\=\int^(2\pi)_0\int_0^(√(8))(r√(9-r^2)-r)\,dr\,d\theta\\\\=\int^(2\pi)_0d\theta\,\cdot\biggr[\int^(√(8))_0r√(9-r^2)\,dr-\int^(√(8))_0r\,dr\biggr]\\\\=2\pi\biggr[\int^(√(8))_0r√(9-r^2)\,dr-4\biggr]\\\\=2\pi\int_0^(√(8))r√(9-r^2)\,dr-8\pi

Let
u=9-r^2 and
\,du=-2r\,dr. Bounds now become
u_1=9-0^2=9 to
u_2=9-(√(8))^2=1:


\displaystyle -\pi\int_9^1√(u)\,du-8\pi\\\\=\pi\int_1^9√(u)\,du-8\pi\\\\=\pi\biggr((2)/(3)(9)^(3)/(2)-(2)/(3)(1)^(3)/(2)\biggr)-8\pi\\\\=\pi\biggr(18-(2)/(3)\biggr)-8\pi\\\\=(52\pi)/(3)-(24\pi)/(3)\\\\=(28\pi)/(3)

Therefore, the volume of the region is
(28\pi)/(3) cubic units.

User Carlodef
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