Thread: Volume of a solid of revolution

Find the volume of the solid whose base is bounded by the circle x^2+y^2=4 with Equilateral triangle cross sections taken perpendicular to the x-axis.

area of an equilateral triangle of side length " " is

for your problem, the side of an equilateral cross-section is .

equation for the volume of the described solid is



get the cross-sectional area as a function of x and integrate.

I was able to work it out and got V=8. For semicircles, did i work it correctly?
y=(4-x^2)^(1/2)
A(x)=(pi r^2)/2
r=2(4-x^2)^(1/2)

V= Integral pi((4-x^2)^2)dx Limits [-2,2]= 32pi/3

Originally Posted by HMV

I was able to work it out and got V=8.

not correct. how did you get V = 8?

For semicircles, did i work it correctly?
y=(4-x^2)^(1/2)
A(x)=(pi r^2)/2
r=2(4-x^2)^(1/2)

V= Integral pi((4-x^2)^2)dx Limits [-2,2]= 32pi/3

you got double the correct volume ... r = (4-x^2)^(1/2)

note that it's much easier if you realized that it's just the volume of a hemisphere of radius 2 ...

V = (2/3)pi*2^3 = 16pi/3

ok?