using discs (washers),
v = ∫[0,3^(1/5)] π(R^2-r^2) dx
where R=3 and r=y=x^5
v = ∫[0,3^(1/5)] π(9-x^10) dx = 90π/11 3^(1/5)
using shells,
v = ∫[0,3)] 2πrh dy
where r=y and h=x=y^(1/5)
v = ∫[0,3)] 2πy^(6/5) dy = 90π/11 3^(1/5)
You might have showed your work, right?
Find the volume of the solid of revolution obtained by revolving the plane region R bounded by y= x^5, the y-axis, and the line y=3 about the x-axis
I've gotten 15/7(3^(2/5))pi, but it's not right. Please help
1 answer