Question

A solid generated when region in the first quadrant is enclosed by curves y=2x^2 and y^2=4x is revolved around the x-axis. What is the volume?

A. Pi/5
B. 2Pi/5
C. 8Pi/5
D. 6Pi/5

Answer 1

solve for intersections

sub the first eqn into the second equation

4x^4= 4x
x^4 - x =0
x(x^3-1) =0

x=0, x=1 for real solutions

Answer 2

See the picture

this is equal to the volumeof rotation below the curve y^2 = 4x minus the volume of rotation found by rotating the lower curve ( ie y=2x^2)

Remember

Volume of curve about x axis form x=a to x=b is

\[V = \int\limits_{a}^{b} y^2 dx\]

Answer 3

so our volume is equal to \[V= \int\limits_{0}^{1} ( 4x ) dx - \int\limits_{0}^{1} ( 2x^2)^2 dx\]

Answer 4

\[ V= \int\limits_{0}^{1} 4x - 4x^4 dx \]

Answer 5

Whoops, there should be a actor of pi outside the the integrals

Answer 6

so answer = pi [ 2 - 4/5] = 6pi/ 5 = \[\frac{6\pi}{5}\]