Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y2 = 2x, x = 2y; about the y-axis I'm getting 64pi/15. Can anyone come up with a different answer?

Answer 1

That's the region we are rotating about the y-axis. Note that we will be integrating with respect to y. Note that : y^2 = 2x is the same as: x = (1/2)y^2

We also have x = 2y.

The outer radius is given by: R(y) = 2y - 0 = 2y.

The inner radius is given by: r(y) = (1/2)y^2 .

We also need to find the points of intersection. Equate x = 2y and x = (1/2)y^2

2y = (1/2)y^2 , 4y = y^2, y^2 - 4y = 0, y(y - 4) = 0

y = 0 and y = 4.

So the volume is given by: \(V = \pi \int_{0}^{4} [(2y)^2 -(\frac{y^2}{2})^2]dy\)

\(V = \pi \int_{0}^{4} [4y^2 - \frac{y^4}{4}]dy\) = \(\pi [\frac{4y^3}{3} - \frac{y^5}{20}] |_{0}^{4}\)

= \(\pi *[\frac{256}{3} - \frac{256}{5}] =\pi *[\frac{256*5 - 256*3}{15}] = \frac{512\pi}{15}\)

Which is your final answer.

Answer 2

Region of rotation 2 10 12 .2

Consi der the region of rotatiorn Region is rotated about y axis for the point of intersection, -4)0 The region lies in 0sys4 Cross secti on area of washer is A (y) = π(Outer radius). π (inner radius )2 4 volume of the solid of revolution is 4 (64)- (1024)} π(25.120) 3 20 5120 3072 60 2048 60 512