Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
[Calculus] Volume of Shapes ?
1) Find volume of y=sqrt(9-x^2), with bounds y=0, x=0, rotated around x=4
2) Find volume of x^(2/3), with bounds x=1, y=0, rotated around the y-axis
3) Find volume of y=(1/2)x+1, with bounds y=1, x=0, x=4, rotated around x=4
1 Answer
- az_lenderLv 71 year agoFavorite Answer
(1) Note that this is a part of the circle x^2+y^2=9. The question is defective, because it does not say whether x=0 is the left boundary or the right boundary. (Your teacher is a moron.) I'll assume it's the left boundary. Then the volume is
Integral from x = 0 to 3 of
2*pi*r*h dx
= integral of 2*pi*(4 - x)*sqrt(9-x^2) dx.
The integral of x*sqrt(9-x^2) is easy by substituting u = 9-x^2.
The integral of 4*sqrt(9-x^2) should be done by substituting x = 3*sin(theta).
I leave you to finish the problem.
(2) "Volume of x^(2/3)" -- I guess your moron teacher means that the curve y = x^(2/3) is one of the boundaries. Here again we need the "shells" method. Volume is
Integral from x = 0 to 1 of
2*pi*r*h dx
= integral of 2*pi*x*x^(2/3) dx
= integral of 2*pi*x^(5/3) dx.
After integration you have
(3/4)*pi*x^(8/3), and plugging in the limits gives
(3/4)*pi.
(3) This one is just a cone. The radius is 4, the height is 3-1 = 2. The volume is
(1/3)pi*16*2 = (32*pi)/3.