Question

Find the volume of the solid obtained by rotating the region bounded by y=27x^3, y=0, x=1 about x=2

Answer 1

Hello,

$V=\int_{0}^{27}\pi(2-x)^2dy\\\\ =\pi*\int_0^1(2-x)^2*81x^2dx\ since\ dy=27*3*x^2*dx\\\\ =\pi* \left[ \dfrac{4x^3}{3}- \dfrac{x^4}{2}+ \dfrac{x^5}{5} \right]_0^1\\\\ = \dfrac{31\pi}{30}$

Answer 2

The volume of the solid obtained by rotating the region is $\mathbf{\lmathbf{ \dfrac{81 \pi}{5}}}$

The volume of the solid obtained by rotating the region bounded by the curve can be determined by using the cylindrical shell method.

The cylindrical shell method can be computed by using the formula:

$\mathbf{V = \int ^x_y 2 \pi rh \ dx}$

The sketch for the curves y = 27x³ ; y = 0 ; x = 1; about x = 2 can be seen in the image attached below.

Using the information given in the sketch below:

• The radius of the shell = (2-x)
• The circumference of the shell = 2πr = 2π(2-x)
• The height of the shell = 27x³

Thus, the volume of the solid obtained by using the cylindrical shell method can be calculated as:

$\mathbf{V = \int ^x_y 2 \pi rh \ dx}$

$\mathbf{V = \int ^1_0 2 \pi (2-x) (27x^3) \ dx}$

$\mathbf{\lmathbf{\implies 54 \pi \int^1_0 (2x^3-x^4) \ dx}}$

$\mathbf{\lmathbf{\implies 54 \pi\Big (\dfrac{2x^4}{4}-\dfrac{x^5}{5} \Big)^1_0}}$

$\mathbf{\lmathbf{\implies 54 \pi\Big (\dfrac{2(1)^4}{4}-\dfrac{(1)^5}{5} -0\Big)}}$

$\mathbf{\lmathbf{\implies 54 \pi\Big (\dfrac{1}{2}-\dfrac{1}{5}\Big)}}$

$\mathbf{\lmathbf{\implies 54 \pi\Big (\dfrac{3}{10}\Big)}}$

$\mathbf{\lmathbf{\implies \dfrac{81 \pi}{5}}}$

Therefore, we can conclude that the volume of the solid is $\mathbf{\lmathbf{ \dfrac{81 \pi}{5}}}$

Learn more about the cylindrical shell method here:

https://brainly.com/question/13497751