Let f and g be the functions given by f(x)=1+sin(2x) and g(x)=e^(x/2). Let R be the shaded region in the first quadrant enclosed by the graphs of f and g.

Question

A. Find the the area of R.
B. Find the value of z so that x=z cuts the solid R into two parts with equal area.
C. Find the volume of slid generated when R is revolved around the x-axis.

Answer 1

I would start by sketching it
http://www.wolframalpha.com/input/?i=plot+y+%3D+1+%2B+sin%282x%29%2C+y+%3D+e%5E%28x%2F2%29

So now for the hard part of finding their intersection.
The left one is obviously (0,1)
How do they expect you to solve
1 + sin(2x) = e^(x/2) ??, Have you learned any methods such as Newton's Method ??

Wolfram says:
http://www.wolframalpha.com/input/?i=solve++1+%2B+sin%282x%29+%3D+e%5E%28x%2F2%29

x = 1.13569..
(I tested it , it works)

so now you have to integrate
1 + sin(2x) - e^(x/2) from x = 0 to x = 1.13569

The integration itself should pose no problem, just a lot of buttonpushing after that.
So that would be part A

Man, this is a major question!
This looks like a major assignment, and I just can't justify giving you the solution.
I gave you a good start, so hang in there.

hint for B.
you have to set the two integrals from 0 to z equal to (1.13569 - z) and solve for z

hint for C
use "washers" , that is the outer radius is the 1+sin2x
the inner radius is the e^(x/2) curve

Answer 2

I got .429 u^2 for A.

I don't understand your hint for B..

Answer 3

For C i for 1.348 u^3. Is that right?