Answer
$\dfrac{224\pi}{15}$
Work Step by Step
Our aim is to compute the volume of the revolution of the curve about the y-axis by using the Washer method.
Washer method for computing the volume of the revolution of the curve: Let us consider two functions $f(x)$ and $g(x)$ (both are continuous functions) with $f(x) \geq g(x) \geq 0$ on the interval $[m, n]$ . Then the volume of the solid can be obtained by rotating the region under the graph about the y-axis and can be expressed as: $\ Volume, V=\pi \int_m^n [f(y)^2-g(y)^2] \ dy $
$\bf{Calculations:}$
$V= \pi \int_0^2 [(4)^2-(4-y^{2})^2] \ dy$
or, $= \pi \int_0^2 (8y^2-y^4) \ dy$
or, $= \pi \times ( \dfrac{8y^{3}}{3}-\dfrac{y^5}{5})_0^2$
or, $=\pi \times ( \dfrac{8(2)^{3}}{3}-\dfrac{(2)^5}{5})$
Hence, $V=\dfrac{224\pi}{15}$