## University Calculus: Early Transcendentals (3rd Edition)

$$\dfrac{ 4 \pi\space r^3}{3}$$
Consider the equation of the circle about the x- axis $r^2=x^2+y^2$ Consider the washer method to compute the volume: $$V=(2) \int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dy \\= \int_{0}^{r} 2y (\sqrt {r^2-y^2}) \space dy \\=[\dfrac{-4 \pi \times (r^2-y^2)^{(3/2)}}{3}]_{0}^{r} \\=\dfrac{ 4 \space \pi\space r^3}{3}$$