Find the volumes of the solids obtained by rotating the region bounded by the curves y = x and y = x^2 about the following lines.(a) The x-axis(b) The y-axis(c) y = 2

(a) A cross-section is a washer with inner radius [tex]x^2[/tex] and outer radius [tex]x[/tex].

[tex]V = \int_0^1 \pi\left[(x)^2 - (x^2)^2\right]dx = \int_0^1 \pi(x^2 - x^4) dx = \pi \left[ \frac{1}{3}x^3 - \frac{1}{5}x^5\right]_0^1 \\ \\ = \pi \left[\frac{1}{3} - \frac{1}{5}\right] = \frac{2}{15}\pi[/tex]

(b)A cross-section is a washer with inner radius [tex]y[/tex] and outer radius [tex]\sqrt{y}[/tex].

[tex]V = \int_0^1 \pi \left[ \left( \sqrt{y} \right)^2 - y^2 \right]dy = \int_0^1 \pi(y - y^2)dy = \pi \left[ \frac{1}{2}y^2 - \frac{1}{3}y^3\right]_0^1 \\ = \pi \left[ \frac{1}{2} - \frac{1}{3} \right] = \frac{\pi}{6}[/tex]

(c) A cross-section is a washer with inner radius [tex]2-x[/tex] and outer radius [tex]2 - x^2[/tex]

[tex]V = \int_0^1 \pi\left[ (2-x^2)^2 - (2-x)^2\right] = \int_0^1 \pi (x^4 - 5x^2 + 4x) dx \\ \\ = \pi \left[ \frac{1}{5}x^5 - \frac{5}{3}x^3 + 2x^2 \right]_0^1 = \pi\left[ \frac{1}{5} - \frac{5}{3} + 2\right] = \frac{8}{15}\pi[/tex]