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$\displaystyle \pi$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$\displaystyle \pi$(1.5)^2(3) = 2.25$\displaystyle \pi$
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$\displaystyle \pi$
Area of the Ellipse = (4/3)$\displaystyle \pi$(1.73)(3)(1.73) = ((4/3)(9)$\displaystyle \pi$)/2 <--- remember its only half a ellipse
so the answer is 6$\displaystyle \pi$ - 2.25$\displaystyle \pi$ = 3.75$\displaystyle \pi$... i think.. O and don't forget to include unit^3 in there.