# Thread: Volume of a solid of revolution

1. ## 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.

2. area of an equilateral triangle of side length "$\displaystyle s$" is $\displaystyle A = \frac{\sqrt{3}}{4} s^2$

for your problem, the side of an equilateral cross-section is $\displaystyle s = 2y$.

equation for the volume of the described solid is

$\displaystyle V = \int_{-2}^{2} A(x) \, dx$

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

3. 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

4. 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?