Answer to Question #121648 in Calculus for Rahul

Question

Find the volume of the solid that results when the region enclosed by

Answer 1

Let solid is obtained by $y=\sqrt x$ , x-axis and revolve it about $x=9$

then Volume, $V = \int_0^9 \pi r^2 dx = \int_0^9 \pi (\sqrt x)^2 dx = \int_0^9 \pi x dx$

$V = [\frac{\pi}{2}(x^2) ]_0^9 = \frac{\pi}{2}(81-0) = \frac{81 \pi}{2}$ unit volume.

Region is bounded by red, blue and green curve is required region.