Question

Find the volume of the solid that is obtained when the region under the curve y= 3sqrt(x+3) over the interval [5, 24] is revolved about x-axis.

Answer 1

एबहप

Answer 2

cross-sections are circular discs
volume is sum of circular areas with radius of f(x)
\[V = \pi \int\limits_{5}^{24}(3\sqrt{x+3})^{2} dx\]

Answer 3

The answer is 3pi/5(3^5 - 2^5) = 633 pi / 5