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nagasenaz

Answered question

2021-08-07

Use polar coordinates to find the volume of the given solid. Inside both the cylinder $x^{2} + y^{2} = 4 and the elepsoid 4 x^{2} + 4 y^{2} + z^{2} = 64 .$

Answer & Explanation

Roosevelt Houghton

Skilled2021-08-08Added 106 answers

Consider the cylinder,

x^{2} + y^{2} = 4

and the ellipsoid,

4 x^{2} + 4 y^{2} + z^{2} = 64

In polar coordinates, we know that

x^{2} + y^{2} = r^{2}

so the ellipsoid gives:

z^{2} = 64 - 4 r^{2}

\Rightarrow z = \pm \sqrt{64 - 4 r^{2}}

So, the volume of the solid is given by V=

\int_{0}^{2 π} \int_{0}^{2} (\sqrt{64 - 4 r^{2}} - (- \sqrt{64 - 4 r^{2}})) r d r d θ

2 \int_{0}^{2 π} \int_{0}^{2} r \sqrt{64 - 4 r^{2}} r d r d θ

to solve this integral, we substitute,

64 - 4 r^{2} = t - 8 r d r = d t \Rightarrow r d t = \frac{- 1}{8} d t

hence the indefinite integral is

\int r \sqrt{64 - 4 r^{2}} = \int \sqrt{t} \cdot \frac{- 1}{8} d t = - \frac{1}{8} \cdot \frac{2}{3} t^{\frac{3}{2}} = - \frac{1}{12} {(64 - 4 r^{2})}^{\frac{3}{2}}

so on applying the limit, the volume becomes

V = 2 \int_{0}^{2 π} \int_{0}^{2} r \sqrt{64 - 4 r^{2}} r d r d θ

2 \int_{0}^{2 π} {[- \frac{1}{12} {(64 - 4 r^{2})}^{\frac{2}{3}}]}_{0}^{2} d θ

= - \frac{1}{6} \int_{0}^{2 π} [{(64 - 4 {(2)}^{2})}^{\frac{3}{2}} - {(64 - 4 {(0)}^{2})}^{\frac{3}{2}}] d θ

= - \frac{1}{6} \int_{0}^{2 π} [{(48)}^{\frac{3}{2}}

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