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
az_lender
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(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.