Calculus 2 rotation question

Question

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

y=18x-2x^2

y=0

about the y-axis

Answer 1

This function is an inverted parabola with a vertex at (4.5, 40.5) and x-intercepts at 0 and 9. A rotation around the y-axis will result in a solid where the cross section is a "washer" shape and the volume can be calculated by integrating that washer are from the x-axis to the top of the shape (the vertex)

The area of the washer is π(R_o² - R_i²) where R_o and R_i are x-values. Unfortunately, we have function written as y(x) instead of x(y) so we need to flop it around to make this work.

y = -2x² + 18x

y = -2(x² - 9x )

y - 81/2 = -2(x² - 9x + 81/4)

y - 40.5 = -2(x - 9/2)²

-0.5y + 20.25 = (x - 4.5)²

±√(-0.5y + 20.25) = x - 4.5

x = 4.5 ± √(-0.5y + 20.25)

Doing a little checking will convince you that:

R_o = 4.5 + √(-0.5y + 20.25)

R_i = 4.5 - √(-0.5y + 20.25)

Now, we need to square these.

R_o² = [4.5 + √(-0.5y + 20.25)]²

R_o² = 4.5² + 9√(-0.5y + 20.25) + (√(-0.5y + 20.25))²

R_o² = 20.25 + 9√(-0.5y + 20.25) + (-0.5y + 20.25)

R_o² = 40.5 - 0.5y + 9√(-0.5y + 20.25)

R_i² = [4.5 - √(-0.5y + 20.25)]²

R_i² = 4.5² - 9√(-0.5y + 20.25) + (√(-0.5y + 20.25))²

R_i² = 20.25 - 9√(-0.5y + 20.25) + (-0.5y + 20.25)

R_i² = 40.5 - 0.5y - 9√(-0.5y + 20.25)

And then subtracting:

R_o² - R_i² = [40.5 - 0.5y + 9√(-0.5y + 20.25)] - [40.5 - 0.5y - 9√(-0.5y + 20.25)]

R_o² - R_i² = 18√(-0.5y + 20.25)

So now we are ready to integrate:

V = ₀∫^40.5(π(18√(-0.5y + 20.25)dy

V = 18π ₀∫^40.5(√(-0.5y + 20.25)dy

V = 18π[-4/3(-1/2y + 81/4)^3/2]₀^40.5

V = 18π[-4/3(-1/2(40.5) + 81/4)^3/2 - -4/3(-1/2(0) + 81/4)^3/2]

V = 18π[0 + 4/3((81/4)^1/2)³]

V = 18π[4/3((9/2)³]

V = 18π[4/3(729/8)]

V = 18π(729/6)

V = 3π(729)

V = 2187π