Find the volume of the solid generated by revolving the region about the given line.?

Question

The region in the second quadrant bounded above by the curve y = 4 - x2, below by the x-axis, and on the right by the y-axis, about the line x = 1
a. (32/3)π b. (56/3)π c. (256/15)π d. 8π

Answer 1

Calculate the volume of revolution about the line x = 1.

Use the method of cylindrical shells.
Evaluate from x = -2 to x = 0.

V = ∫2πry dx = 2π∫[(1 - x)(4 - x²)] dx

= 2π∫[4 - 4x - x² + x³] dx

= 2π[4x - 2x² - x³/3 + x^4/4] | [Evaluated from -2 to 0]

= 0 - 2π[-8 - 8 + 8/3 + 4] = 2π(28/3) = 56π/3

The answer is b. (56/3)π.

Answer 2

This requires integration, using the shell method.

2π ∫ [-2,0] x (-x^2+4) dx

Unfortunately, I haven't learned enough calculus to evaluate that. But, it is the answer, given in Cartesian square units.

I was assuming that "x2" meant "x^2".

I'm not sure what you mean by the a. b. c. d. bits at the bottom. So I ignored them.