Question

Compute the volume for the solid of revolution generated by rotating the region
bounded by the curve y = x3 + 1, x-axis and the lines x = 3, y = 9 around the line
x = 3.

Answer 1

wat?

Answer 2

I know how to compute the area of the region but what next?

Answer 3

the curve is x^3 + 1 not x3 +1. sorry

Answer 4

so we got something like this right?

Answer 5

since we are rotating around x=3; we can use the shell method easily to find the volume

Answer 6

(x+3)^3 +1 = 0

(x+3)^3 = -1

x+3 = -1

x = -4 ... our integration will be from -4 to 0 then

Answer 7

\[2pi\int_{-4}^{0}x[f(x)]dx\]

\[2pi\int_{-4}^{0}x[(x+3)^3+1]dx\]

expand it and do the power rules

Answer 8

check that y=9 part tho,

9 = (x)^3 +1

8 = (x)^3

2 = x when x = -1 ... so thats a part i dint account for :)

Answer 9

2-3= -1; so integrate that first part from -4 to -1; and add the cylindar volume formed by a radius = 1 height = 9