The circumference of the shell is c(x) = 2πx
The height of the shell is [tex]h(x) = 5x(x-2)^{2}[/tex]
The required volume of the shell is [tex]\frac{32\pi}{3}[/tex].
Given that,
S be the solid obtained by rotating the region,
Where, a = 5 and b =2
We have to determine ,
What are its circumference c and height h.
According to the question,
- The volume is the summation of the volumes of the shells. The volume of a shell is its circumference ,
c(x) = 2πx
- And the height of the shell.
[tex]h(x) = 5x(x-2)^{2}[/tex]
- The volume of the shell is given by,
[tex]V = \int (circumference) \ )(height )\ . dx\\\\[/tex]
Where dx thickness of the shell.
Substitute the value in the equation,
[tex]v = \int\limits^2_0 {2\pi x .(5x) (x-2)^{2} }\, dx\\\\v = \int\limits^2_0 {10\pi x^{2} (x-2)^{^{2}} \ d x\\\\\\v =10\pi \int\limits^2_0 {x^{2}.(x^2+4-4x)} \, dx \\\\V = 10\pi \int\limits^2_0( {x^{4} + 4x^{2} - 4x^{3}) \, dx[/tex]
[tex]v = 10\pi \ [ \dfrac{x^{5}}{5} + \dfrac{4x^{3}}{3} - x^{4}]^{2}_0\\\\v = 10\pi [ \dfrac{2^{5}}{5} + \dfrac{4.2^{3}}{3} - 2^{4} - \dfrac{0^{5}}{5} -\dfrac{4.0^{3}}{3} + 0^{4}]}\\\\v = 10\pi \ [ \dfrac{32}{5} + \dfrac{32}{3} - 16 -0-0+0]\\\\v = 10\pi [\dfrac{96+160-240}{15}]\\\\v = 10\pi [\dfrac{16}{15}]\\\\v = \dfrac{32\pi }{3}[/tex]
Hence, The required volume of the shell is [tex]\dfrac{32\pi}{3}[/tex].
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