What integral will find the volume of the solid that is formed when the region bounded by the graphs of y=e^x, x=1, and y=1 is revolved around the line y=-2?

Question

What integral will find the volume of the solid that is formed when the region bounded by the graphs of y=e^x, x=1, and y=1 is revolved around the line y=-2?

Answer 1

Let me offer a comparison.... the integral ((ex+2)2-9)dx from 0 to 1 will give this result in the integral

(e2x+4ex+4-9)dx = (e2x+4ex-5)dx = (.5e2x+4ex-5x)01 = 5.068

I believe the answer is 15.92 Using the Pappus Method

First we find the area of the region which is easily found to be .7183

Then find the y center of area of the region which I calculate as 1.528

The length of the "arm" which you will revolve is 1.528 + 2 = 3.528 (-2 below the x axis and 1.528 above the x axis and let's call this result R) Then using the Pappus Formula

Volume of Revolution = 2*pi*R*Area = 2*pi*3.528*.7183 = 15.92

Answer 2

I got V=∫₀¹⁽(e^x+2)²-9)dx

Make a careful sketch will help to define the region

Jim