volumes (washer/disk method)

Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.

y = x²,  y = 4x; about the y-axis

I tried solving this question and I checked my work multiple times but in my homework website the answer is not being accepted. I got the answer (2048pi)/15. May I please ask where I could have gone wrong?

Answer 1

First, express x in terms of y:

x = √y

x = y/4

Intersection points are x = 0, 4. When x = 0, 4, y = 0, 16, which is the region of integration we'll use.

Because we're rotating about the y-axis, the right curve is the outer, and the left curve is the inner. This leads to the following integral:

After evaluating, you get 128π/3

Hope this was helpful.

Answer 2

The curves intersect at the points (0,0) and (4, 16).

Take a thin horizontal cross section of the region of height y and thickness Δy.

If y = 4x, then x = (1/4)y. If y = x², then x = √y.

Rotating the cross section about the y-axis yields a washer with outer radius √y and inner radius (1/4)y.

Volume of cross section = π[(√y)² - (y/4)²]Δy= π(y - y²/16)Δy

Volume of solid = π ∫_{(0 to 16)} [ y - y²/16 ]dy = 128π/3

Answer 3

x² = 4x; x = 0 or x = 4.

By sholder method volume v = 2π∫₀⁴x(4x - x²)dx = 2π(4x³/3 - x⁴/4)₀⁴ = 128π/3

By washer/disk method: v = π∫₀¹⁶(y - y²/16)dy = π(y²/2 - y³/48)₀¹⁶ = π(128 - 256/3) = 128π/3