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The Volume of the given solid using polar coordinate is:[tex]\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta[/tex]
V= [tex]\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta[/tex]
What is Volume of Solid in polar coordinates?
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,[tex]x^{2}+y^{2} =1[/tex] and the ellipsoid, [tex]4x^{2}+ 4y^{2} + z^{2} =64[/tex]
In polar coordinates, we know that
[tex]x^{2}+y^{2} =r^{2}[/tex]
So, the ellipsoid gives
[tex]4{(x^{2}+ y^{2)} + z^{2} =64[/tex]
4([tex]r^{2}[/tex]) + [tex]z^{2}[/tex] = 64
[tex]z^{2}[/tex] = 64- 4([tex]r^{2}[/tex])
z=± [tex]\sqrt{64-4r^{2} }[/tex]
So, the volume of the solid is given by:
V= [tex]\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta[/tex]
= [tex]2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta[/tex]
To solve the integral take, [tex]64-4r^{2}[/tex] = t
dt= -8rdr
rdr = [tex]\frac{-1}{8} dt[/tex]
So, the integral [tex]\int\ r\sqrt{64-4r^{2} } rdr[/tex] become
=[tex]\int\ \sqrt{t } \frac{-1}{8} dt[/tex]
= [tex]\frac{-1}{12} t^{3/2}[/tex]
=[tex]\frac{-1}{12} (64-4r^{2}) ^{3/2}[/tex]
so on applying the limit, the volume becomes
V= [tex]2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta[/tex]
=[tex]\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta[/tex]
V = [tex]\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta[/tex]
Since, further the integral isn't having any term of [tex]\theta[/tex].
we will end here.
The Volume of the given solid using polar coordinate is:[tex]\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta[/tex]
Learn more about Volume in polar coordinate here:
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