Nevermind, I figured it out
V=∫[0,3]2pix(45-x^2-4x^2)dx
Find the volume of the solid generated by revolving the region in the first quadrant that is above the parabola y= 4x^2 and below the parabola y= 45-x^2 about the y-axis
I just need help setting up the integral. V=∫[0,3]2pix(45-x^2)dx
What am I doing wrong?
3 answers
where do they hit?
4 x^2 = 45 - x^2
x^2 = 9
x = +/- 3
cylinder height = 45-x^2 - 4 x^2
= 45 - 5 x^2
cylinder thickness = dx
cylinder circumference = 2 pi x
so
V=∫[0,3]2 pi x(45-5x^2)dx
4 x^2 = 45 - x^2
x^2 = 9
x = +/- 3
cylinder height = 45-x^2 - 4 x^2
= 45 - 5 x^2
cylinder thickness = dx
cylinder circumference = 2 pi x
so
V=∫[0,3]2 pi x(45-5x^2)dx
Yes, of course -4-1 = -5 :)
0 answers