User:SoberEmu/Sandbox

$$\mathrm{Let} \; R = f_{1}(y) = \sqrt{\frac{y}{4}}$$

$$\mathrm{Let} \; r = f_{2}(y) = \frac{y}{4}$$

$$V=\pi\int_a^b \left(R^2 - r^2\right)dy$$

$$=\pi\int_0^4 \left[\left(\sqrt{\frac{y}{4}}\right)^2 - \left(\frac{y}{4}\right)^2\right]dy$$

$$=\pi\int_0^4 \left(\frac{y}{4} - \frac{y^2}{16}\right)dy$$

$$=\pi \left[\frac{y^2}{8} - \frac{y^3}{48}\right]_{0}^{4}$$

$$=\pi \left[\frac{4^2}{8}-\frac{4^3}{48} - \left(\frac{0^2}{8}-\frac{0^3}{48}\right)\right]$$

$$=\pi \left(2- \frac{64}{48}\right)$$

$$=\frac{2\pi}{3}\; \mathrm{units}^2$$

$$\approx 2.094 \; \mathrm{units}^2$$