Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. x = 4y^2, y ≥ 0, x = 4; about y = 2

Question

I tried solving this question by simplifying:

integrate (from 0 to 4) 2pi (2- (√x /2))• x dx and I got 96pi / 5 but the professor marked it wrong so I wanted to see what is the right way to solve it and what will be the answer? Thank you in advance.

Answer 1

Shells Method

The radius of the shell should be,

r = 2-y.

This puts the integral in the y-direction. This gives the thickness of the shell as,

dr = dy.

The height of the shell should be

h = 4-x = 4 - 4y².

The intersection of the curve and boundary gives

0 ≤ y ≤ 1.

The volume of the shell should be

dV = 2 π r h dr = 2 π( 2-y)(4 - 4y²)dy

V = 2 π ∫₀¹( 2-y)(4 - 4y²)dy = 26/3 π

Washer Method (just to double check)

Solving for y give the curve as

y = x^1/2/2.

Given the boundary y≥0 we do not need to include the negative part of the curve. Rotating the area round y=2 gives

r_outer = 2-0,

and

r_inner = 2 - x^1/2/2.

The cross-sectional area of the washer should be

A = π(2² - (2 - x^1/2/2)²).

This gives the volume as

V = π∫₀⁴(2² - (2 - x^1/2/2)²) dx = 26/3 π