Question

Let s be the solid obtained by rotating the region shown in the figure about the y-axis (Assume a = 5 and b = 2)


Sketch a typical approximating shell.


What are its circumference c and height h?


c(x) =


h(x) =


Use shells to find the volume V of S.


V =

Answer 1

Answer:

• see attached for a figure
• c(x) = 2πx
• h(x) = 5x(x -2)^2
• V = 32π/3

Step-by-step explanation:

The volume is the summation of the volumes of the shells. The volume of a shell is its circumference ...

  c(x) = 2πx

multiplied by its height ...

  h(x) = 5x(x -2)^2

and its thickness, dx.

That summation is the integral ...

  $\displaystyle V=\int^2_0 {2\pi x(5x)(x-2)^2} \, dx=10\pi\int^2_0 {(x^2-2x)^2} \, dx=10\pi\left(\dfrac{2^5}{5}-\dfrac{4(2^4)}{4}+\dfrac{4(2^3)}{3}\right)\\\\\boxed{V=\dfrac{32\pi}{3}}$

Answer 2

The circumference of the shell  is c(x) = 2πx

The height of the shell is $h(x) = 5x(x-2)^{2}$

The required volume of the shell is $\frac{32\pi}{3}$.

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.

          $h(x) = 5x(x-2)^{2}$

• The volume of the shell is given by,

$V = \int (circumference) \ )(height )\ . dx$

Where dx thickness of the shell.

Substitute the value in the equation,

$v = \int\limits^2_0 {2\pi x .(5x) (x-2)^{2} }\, dx\\\\v = \int\limits^2_0 {10\pi x^{2} (x-2)^{^{2}} \ d x\\\\\\v =10\pi \int\limits^2_0 {x^{2}.(x^2+4-4x)} \, dx \\\\V = 10\pi \int\limits^2_0( {x^{4} + 4x^{2} - 4x^{3}) \, dx$

$v = 10\pi \ [ \dfrac{x^{5}}{5} + \dfrac{4x^{3}}{3} - x^{4}]^{2}_0\\\\v = 10\pi [ \dfrac{2^{5}}{5} + \dfrac{4.2^{3}}{3} - 2^{4} - \dfrac{0^{5}}{5} -\dfrac{4.0^{3}}{3} + 0^{4}]}\\\\v = 10\pi \ [ \dfrac{32}{5} + \dfrac{32}{3} - 16 -0-0+0]\\\\v = 10\pi [\dfrac{96+160-240}{15}]\\\\v = 10\pi [\dfrac{16}{15}]\\\\v = \dfrac{32\pi }{3}$

Hence, The required volume of the shell is $\dfrac{32\pi}{3}$.

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