Question

The shaded region R in diagram below is enclosed by y-axis, y = x^2 - 1 and y = 3.

Answer 1

Cross sections of the volume are washers or annuli with outer radii x(y) + 1, where

y = x(y) ² - 1   ==>   x(y) = √(y + 1)

and inner radii 1. The distance between the outermost edge of each shell to the axis of revolution is then 1 + √(y + 1), and the distance between the innermost edge of R on the y-axis to the axis of revolution is 1.

For each value of y in the interval [-1, 3], the corresponding cross section has an area of

π (1 + √(y + 1))² - π (1)² = π (2√(y + 1) + y + 1)

Then the volume of the solid is the integral of this area over [-1, 3]:

$\displaystyle\int_{-1}^3\pi y\,\mathrm dy = \frac{\pi y^2}2\bigg|_{-1}^3 = \boxed{4\pi}$

$\displaystyle\int_{-1}^3 \pi\left(2\sqrt{y+1}+y+1\right)\,\mathrm dy = \pi\left(\frac43(y+1)^{3/2}+\frac{y^2}2+y\right)\bigg|_{-1}^3 = \boxed{\frac{56\pi}3}$