# Find the volume of the resulting solid which is enclosed by the curve y = x and y = x2 is rotated about the x-axis.

The total volume of solid is

$$V = \int dv = \pi \int_a^b (r_0^2 – r_1^2) dx$$

Then in this case, we have

r0 = x

r1 = x2

a = 0

b = 1

Now,

$$V = \int dv = \pi \int_a^b (r_0^2 – r_1^2) dx$$

$$= \pi \int_0^1 [x^2 – (x^2)^2] dx$$

$$= \pi \int_0^1 [x^2 – x^4] dx$$

$$= \pi \left [ \frac{x^{2+1}}{2+1} – \frac{x^{4+1}}{4+1} \right ]_0^1$$

$$= \pi \left [ \frac{x^3}{3} – \frac{x^5}{5} \right ]_0^1$$

$$= \pi \left [ \frac{5x^3 – 3x^5}{15} \right ]_0^1$$

$$= \frac{\pi}{15} \left \{ [5(1)^3 – 3(1)^5] – [5(0)^3 – 3(0)^5] \right \}$$

$$= \frac{\pi}{15} (5.1 – 3.1)$$

$$= \frac{2\pi}{15}$$