but that's not the answer though. Plus are you able to do it my way?
Q. The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.
y=x^3 and y=sqrt.of(x); about x = 1.
What I did: Rotating about a vertical line along the y-axis.
x^3=sqrt.of(x); x= 0 and 1
x=y^(1/3) and x=y^2.
A(y)=pi(outer radius)^2-pi(inner radius)^2
=pi(1+y^(1/3))^2-pi(1+y^2)^2
V= pi*Integral from 0 to 1 of [(1+(y^1/3))^2-(1+y^2)^2]dy
V= pi*Integral from 0 to 1 of [(1+y^(2/3)+2y^(1/3)-(1+y^4+2y^2)]dy
V= pi[(3y^(2/3))/2 + (6y^(4/3))/4 - (y^4)/4 - (2y^3)/3] from 0 to 1
But after plugging in for 0 and 1, I got 25pi/12. But the ans. is 17pi/30...8(
I wonder why you put 1+y^{1/3}... But since I never worked with solid or rotation other than about the main axes, I would have gone with this:
From what I see, you need to move both curves to the left by one unit.
Hence, the equation of the new cubic becomes:
And that of the square root becomes:
The volume then becomes:
Which gives:
Are you sure the answer is 17pi/30? I'm getting 13pi/30...
Or I missed something...