Question

What is the surface area of the solid created by revolving `f(x) = x^2+e^x , x in [2,4]` around the x axis?

Answer 1

`V=pi1044~~3279.82`

Explanation:

First, take a look at the graph `f(x) = x^2+e^x, x in[2,4]`


I've drawn vertical lines at the endpoints to show where the solid would revolve around the `x`-axis. It looks like the object would ultimately resemble a circular bell.

The formula for find the volume of a shape like this is

`V=piint_("lower")^("upper")[f(x)]^2dx`

In this case, you can plug in all the values and find the integral in a straight forward fashion.

`V=pi int_2^4 [x^2+e^x]^2dx`

`V=pi (int_2^4[x^4+2x^2e^x+e^(2x)]dx)`

`V=pi [1/5x^5+2x^2e^x+4xe^x+2e^(2x)]_2^4`

`V=pi[1/5(4)^5+2(4)^2e^4+4(4)e^x+2e^(2*4)]-pi[1/5(2)^5+2(2)^2e^2+4(2)e^2+2e^(2*2)]`

`V=pi1044~~3279.82`