The Volume of the given solid using polar coordinate is:![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
V= ![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
<h3>
What is Volume of Solid in polar coordinates?</h3>
To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.
Consider the cylinder,
and the ellipsoid, 
In polar coordinates, we know that

So, the ellipsoid gives

4(
) +
= 64
= 64- 4(
)
z=± 
So, the volume of the solid is given by:
V= ![\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B2%5Cpi%7D_%200%20%5Cint%5Climits%5E1_0%7B%7D%20%5C%2C%20%5B%5Csqrt%7B64-4r%5E%7B2%7D%20%7D-%20%28-%5Csqrt%7B64-4r%5E%7B2%7D%20%7D%29%5D%20r%20dr%20d%5Ctheta)
= 
To solve the integral take,
= t
dt= -8rdr
rdr = 
So, the integral
become
=
= 
=
so on applying the limit, the volume becomes
V= 
=![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2864-4%281%29%5E%7B2%7D%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864-4%282%29%5E%7B0%7D%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
V = ![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
Since, further the integral isn't having any term of
.
we will end here.
The Volume of the given solid using polar coordinate is:![\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B6%7D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_%20%7B0%7D%20%5B%2860%29%20%5E%7B3%2F2%7D%20%5C%3B%20-%2864%29%20%5E%7B3%2F2%7D%20%5D%20d%5Ctheta)
Learn more about Volume in polar coordinate here:
brainly.com/question/25172004
#SPJ4