Question

Can you please check to see if these answers are correct? I have provided my work as well! That you! This is much appreciated!!

1. What is the volume of the solid that is generated by revolving the region bounded by the curves x = 3y^2-2 and x = y^2 and y = 0 about the x axis?

Shell Method: h = y
r = y^2-(3y^2-2)
My Work: 2pi∫ [0,1] (y)(y^2-3y^2+2)dy
2pi((-2(1)^4/4)+2(1)^2/2)
Answer = 3pi

2. What is the volume of the solid that is generated by revolving the region bounded by the curves x = 3y^2-2 and x = y^2 and y = 0 about the vertical line x = 1?
Washer Method: R = 1 - (3y^2-2)
r = 1-y^2
My Work: pi∫ [0,1] (-3y^2+3)^2-(-1+y^2)^2 dy
pi (8(1)^5/5 - 16(1)^3/3 + 8(1))
Answer = 64pi/15

Answer 1

#1. Using horizontal shells of thickness dy,
r = y
h = y^2 - (3y^2-2) = 2-2y^2

v = ∫[0,1] 2πrh dy
= 2π∫[0,1] y(2-2y^2) dy = π
You got r and h reversed, but the product was luckily the same. Unfortunately, you added instead of subtracting.

Using discs of radius y and thickness dx,
v = ∫π(R^2-r^2) dx
v = ∫[-2,0] π(x+2)/3 dx + ∫[0,1] π((x+2)/3 - x) dx = 2π/3 + π/3 = π

#2. looks good.
Just as a check, using shells of thickness dx,
v = ∫2πrh dx
= ∫[-2,0] 2π(1-x)√((x+2)/3) dx
+ ∫[0,1] 2π(1-x)(√((x+2)/3)-(√x)) dx
= 8π/5 √6 + (64/15 π - 8π/5 √6)
= 64/15 π

Answer 2

Thank you Steve!