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.

D. Find the volume of the solid generated when R is revolved around the line x=pi

E. Find the volume of the solid generated when R is revolved around the line y=pi.

I already asked these questions but have answered some and want to check my answers and i still have questions on some.

For A: A=integral from 0 to 1.136 of (1+sin(2x)-e^(x/2))dx = .429 u^2

For B: I got a hint that said "you have to set the two integrals from 0 to z equal to (1.136-z) and solve for z." I still don't understand this hint.

For C: V=pi times the integral from 0 to 1.136 of [(1+sin(2x))^2-(e^(x/2))^2]dx = 4.267 u^3

For D: V=pi times the integral from 0 to 1.136 of [(1+sin(2x)-pi)^2-(e^(x/2)-pi)^2]dx = -4.204 u^3. What did I do wrong in this problem? I know it shouldn't be a negative number.

For E. I know it's similar to D but I'm not sure how to figure out the bounds.

Answer 1

so far, I don't see any of your work, just the answers others have provided.

For B, if you have to split the area into two equal parts, and the integration goes from 0 to 1.136, that means that there is some z in that interval which bounds exactly half the area. So, you want

∫[0,z] 1+sin(2x)-e^(x/2) dx = ∫[z,1.136] 1+sin(2x)-e^(x/2) dx

You have the indefinite integral. If you evaluate those two integrals, you will get two expressions involving z. Then just solve for z.

D: You revolved around the line y=π, not x=π. That is for E. See below. For D, you will want to use shells, since that lets us integrate along x. The volume of a shell of thickness dx is 2πrh dx, so

v = ∫[0,1.136] 2π(π-x)(1+sin(2x)-e^(x/2)) dx = 6.886

E: Since π is above the region, your expression should be
∫[0,1.136] π((π-e^(x/2))^2-(π-(1+sin(2x)))^2)dx = 4.2036