Question

Let R be the region in the first quadrant bounded by the curves y = x^3 and y = 2x - x^2. Calculate the following quantities.

Answer 1

(a)

$2x - x^2 \ge x^3 \\ \\ A = \int_0^1(2x - x^2 - x^3) dx = \left[ x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4\right]_0^1 \\ \\ A = 1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$

(b) A cross-section is a washer with inner radius $x^3$ and outer radius $2x - x^2$ so its area is $\pi(2x - x^2)^2 - \pi(x^3)^2$.

$V = \int_0^1 A(x) dx = \int_0^1 \pi \left[ (2x - x^2)^2 - (x^3)^2\right] dx \\ \\ = \int_0^1 \pi( 4x^2 - 4x^3 + x^4 - x^6)dx \\ \\ =\pi \left[\frac{4}{3}x^3 - x^4 + \frac{1}{5}x^5 - \frac{1}{7}x^7 \right]_0^1 = \pi \left(\frac{4}{3} - 1 + \frac{1}{5}- \frac{1}{7} \right) \\ \\ =\frac{41}{105}\pi$

(c) Using the method of cylindrical shells,

$V = \int_0^1 2\pi x(2x - x^2 - x^3)dx = \int_0^1\pi (2x^2 - x^3 - x^4)dx \\ \\ = 2\pi \left[\frac{2}{3}x^3 - \frac{1}{4}x^4 - \frac{1}{5}x^5 \right]_0^1 \\ \\ = 2\pi \left( \frac{2}{3} - \frac{1}{4} - \frac{1}{5}\right) = \frac{13}{30}\pi$