evaluate yyye sx2 1 y2 dv, where e is the region that lies inside the cylinder x2 1 y2 − 16 and between the planes z − 25 and z − 4

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25-08-2022
Mathematics

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evaluate yyye sx2 1 y2 dv, where e is the region that lies inside the cylinder x2 1 y2 − 16 and between the planes z − 25 and z − 4

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okpalawalter8 okpalawalter8 · Answer 1 · 2022-08-29T18:22:39+02:00

The region that lies inside the cylinder x2 1 y2 − 16 and between the planes z − 25 and z − 4 is mathematically given as

[tex]\int\int \int_{E}^{2}+y^{2} \mathrm{dv}=384\pi[/tex] (\int\int \int_{E}^{2}+y^{2} \mathrm{dv}=384\pi)

What is the region that lies inside the cylinder x2 1 y2 − 16 and between the planes z − 25 and z − 4?

Generally, the equation for the volume integral: is mathematically given as

[tex]$\iiint_{E} \sqrt{x^{2}+y^{2}} \mathrm{dv}$[/tex]

where

$E$ is the region within the cylinder

x^{2}+y^{2}=16 between the planes z=-5 and z=4.

These cylindrical coordinates will be used for the volume integral calculation that has to be done. The difference in volume when using cylindrical coordinates is calculated as follows:

[tex]\mathrm{dv} = {rdzdrd} \theta[/tex].

The z limits are given as -5 <= z <= 4. With cylindrical coordinates, we have:

[tex]x^{2}+y^{2}=r^{2} 0 \leq x^{2}+y^{2}=r^{2} \leq 16 > 0 \leq r \leq 4.[/tex]

The theta limits are :

[tex]$0 \leq \theta \leq 2 \pi$[/tex].

We have [tex]x^{2}+y^{2}=r^{2} \Rightarrow$ $\sqrt{x^{2}+y^{2}}= r[/tex]

The volume integral is what we have next.

[tex]\sqrt{x^{2}+y^{2}}=r[/tex]

We then have the volume integral is:

[tex]&\iiint_{E} \sqrt{x^{2}+y^{2}} \mathrm{dv}= \\\\\&\int_{\theta=0}^{2 \pi} \int_{r-0}^{4} \int_{z--5}^{4} r(r \mathrm{~d} z \mathrm{~d} r \mathrm{~d} \theta) \\\\\&=\int_{0}^{2 \pi} \int_{0}^{4} \int_{-5}^{4} r^{2} \mathrm{~d} z \mathrm{~d} r \mathrm{~d} \theta \\\\&=\int_{0}^{2 \pi} \int_{0}^{4} r^{2}(z)_{-5}^{4} \mathrm{~d} r \mathrm{~d} \theta \\\\[/tex]

[tex]&=\int_{0}^{2 \pi} \int_{0}^{4} r^{2}(4--5) \mathrm{d} r \mathrm{~d} \theta \\\\&=\int_{0}^{2 \pi} \int_{0}^{4} 9 r^{2} \mathrm{~d} r \mathrm{~d} \theta \\\\&=\int_{0}^{2 \pi}\left(3 r^{3}\right)_{0}^{4} \mathrm{~d} \theta \\\\&=\int_{0}^{2 \pi} 3\left(4^{3}-0\right. \\&=\int_{0}^{2 \pi} 3(64) \mathrm{d} \theta \\\\ =192 \int_{0}^{2 \pi} d \theta \\\\\ =192(\theta)_{0}^{2 \pi} \theta[/tex]

[tex]\int\int \int_{E}^{2}+y^{2} \mathrm{dv}=384\pi[/tex]

In conclusion, the region that lies inside the cylinder is

[tex]\int\int \int_{E}^{2}+y^{2} \mathrm{dv}=384\pi[/tex]