Question

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. (tried this question a lot and I am just missing something)(plz help)

x = −3y2 + 9y − 6, x = 0

Answer 1

The expression on the left side describes a parabola. Factorize it to determine where it crosses the y-axis (i.e. the line x = 0) :

-3y² + 9y - 6 = -3 (y² - 3y + 2)

… = -3 (y - 1) (y - 2) = 0

⇒   y = 1   or   y = 2

Also, complete the square to determine the vertex of the parabola:

-3y² + 9y - 6 = -3 (y² - 3y) - 6

… = -3 (y² - 3y + 9/4 - 9/4) - 6

… = -3 (y² - 2•3/2 y + (3/2)²) + 27/4 - 6

… = -3 (y - 3/2)² + 3/4

⇒   vertex at (x, y) = (3/4, 3/2)

I've attached a sketch of the curve along with one of the shells that make up the solid. For some value of x in the interval 0 ≤ x ≤ 3/4, each cylindrical shell has

radius = x

height = y⁺ - y⁻

where y⁺ refers to the half of the parabola above the line y = 3/2, and y⁻ is the lower half. These halves are functions of x that we obtain from its equation by solving for y :

x = -3y² + 9y - 6

x = -3 (y - 3/2)² + 3/4

x - 3/4 = -3 (y - 3/2)²

-x/3 + 1/4 = (y -  3/2)²

± √(1/4 - x/3) = y - 3/2

y = 3/2 ± √(1/4 - x/3)

y⁺ and y⁻ are the solutions with the positive and negative square roots, respectively, so each shell has height

(3/2 + √(1/4 - x/3)) - (3/2 - √(1/4 - x/3)) = 2 √(1/4 - x/3)

Now set up the integral and compute the volume.

$\displaystyle 2\pi \int_{x=0}^{x=3/4} 2x \sqrt{\frac14 - \frac x3} \, dx$

Substitute u = 1/4 - x/3, so x = 3/4 - 3u and dx = -3 du.

$\displaystyle 2\pi \int_{u=1/4-0/3}^{u=1/4-(3/4)/3} 2\left(\frac34 - 3u\right) \sqrt{u} \left(-3 \, du\right)$

$\displaystyle -12\pi \int_{u=1/4}^{u=0} \left(\frac34 - 3u\right) \sqrt{u} \, du$

$\displaystyle 12\pi \int_{u=0}^{u=1/4} \left(\frac34 u^{1/2} - 3u^{3/2}\right)  \, du$

$\displaystyle 12\pi \left(\frac34\cdot\frac23 u^{3/2} - 3\cdot\frac25u^{5/2}\right)  \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 u^{3/2} - \frac65u^{5/2}\right)  \bigg|_{u=0}^{u=1/4}$

$\displaystyle 12\pi \left(\frac12 \left(\frac14\right)^{3/2} - \frac65\left(\frac14\right)^{5/2}\right) - 12\pi (0 - 0)$

$\displaystyle 12\pi \left(\frac1{16} - \frac3{80}\right) = \frac{12\pi}{40} = \boxed{\frac{3\pi}{10}}$