The shaded region R in diagram below is enclosed by y-axis, y = x^2 - 1 and y = 3. Determine the volume of the solid generated when the shaded region R is revolved
about x = -1 by using Disk method.

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]:

i think it is c, not positive though

step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18