# Finding the volume revolved around the given axis?

• Apr 19th 2010, 11:18 AM
AlphaRock
Finding the volume revolved around the given axis?
y = x^2
y = 2x

What is the volume of the region bounded by y = x^2 and y = 2x around y = 3?
• Apr 19th 2010, 12:59 PM
Warrenx
Quote:

Originally Posted by AlphaRock
y = x^2
y = 2x

What is the volume of the region bounded by y = x^2 and y = 2x around y = 3?

I guess the basics steps for solving this would be:

find the area of the half-ellipse [formed by y=x^2 around y=3] = 4/3 $\pi$abc

then subtract the area of the internal cone formed by the revolution of y=2x from the point of origin following the radius of x=3/2.

so the area of the cone is 1/3 $\pi$(1.5)^2(3) = 2.25 $\pi$

now just find the area of the half-ellipse formed by y=x^2 revolved around the y axis with a bound at {y=0 and y=3}.

then subtract:

Area of half an Ellipse - 2.25 $\pi$

Area of the Ellipse = (4/3) $\pi$(1.73)(3)(1.73) = ((4/3)(9) $\pi$)/2 <--- remember its only half a ellipse

so the answer is 6 $\pi$ - 2.25 $\pi$ = 3.75 $\pi$... i think.. O and don't forget to include unit^3 in there.