Question

Find the volume of the solid generated by revolving the region bounded by

Answer 1

Answer:

$V =\dfrac{25\pi}{\sqrt{2}}$

Step-by-step explanation:

given,

y=5√sinx

Volume of the solid by revolving

$V = \int_a^b(\pi y^2)dx$

a and b are the limits of the integrals

now,

$V = \int_a^b(\pi (5\sqrt{sinx})^2)dx$

$V =25\pi \int_{\pi/4}^{\pi/2}sinxdx$

$\int sin x = - cos x$

$V =25\pi [-cos x]_{\pi/4}^{\pi/2}$

$V =25\pi [-cos (\pi/2)+cos(\pi/4)]$

$V =25\pi [0+\dfrac{1}{\sqrt{2}}]$

$V =\dfrac{25\pi}{\sqrt{2}}$

volume of the solid generated is equal to $V =\dfrac{25\pi}{\sqrt{2}}$