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.
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.
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
For part A:
To find the area of the shaded region R, you need to find the difference between the two curves f(x) and g(x) and integrate over the appropriate interval. This can be expressed as the integral of (f(x) - g(x)) dx from 0 to the point(s) of intersection.
In this case, you correctly found the integral as A = ∫[0 to 1.136] (1 + sin(2x) - e^(x/2)) dx, which evaluates to approximately 0.429 square units. So your answer for part A is correct.
For part B:
To find the value of z at which the solid R is divided into two parts of equal area, you need to set up an equation and solve for z. The equation is:
∫[0 to z] (1 + sin(2x) - e^(x/2)) dx = ∫[z to 1.136] (1 + sin(2x) - e^(x/2)) dx,
since the two parts have equal areas. By setting these two integrals equal to each other, you can solve for z.
For part C:
To find the volume of the solid generated when R is revolved around the x-axis, you need to use the method of cylindrical shells. The volume can be calculated using the formula:
V = π * ∫[0 to 1.136] [(f(x))^2 - (g(x))^2] dx.
In your calculation, you correctly evaluated the integral as V = π * ∫[0 to 1.136] [(1 + sin(2x))^2 - (e^(x/2))^2] dx, which evaluates to approximately 4.267 cubic units. So your answer for part C is correct.
For part D:
To find the volume of the solid generated when R is revolved around the line x = π, you need to use the method of cylindrical shells again. However, in this case, the axis of rotation is different.
The correct formula to use is:
V = π * ∫[0 to 1.136] [(f(x) - π)^2 - (g(x) - π)^2] dx.
By plugging in the values of f(x), g(x), and the axis of rotation into the formula, you should get the correct volume. Double-check your calculations to see if you made any mistakes.
For part E:
To find the volume of the solid generated when R is revolved around the line y = π, you need to change your approach. Instead of using cylindrical shells, you can use the method of disks (or washers).
The formula to use is:
V = π * ∫[0 to 1.136] [(f(x))^2 - (g(x))^2] dy.
Since the axis of rotation is y = π, you need to express the area elements in terms of dy instead of dx. Make sure to adjust your integral limits accordingly based on the bounds mentioned in the problem statement.
For part A, your calculation is correct. The area of the shaded region R is approximately 0.429 square units.
For part B, the hint is suggesting that you can find the value of z by setting the integrals of f(x) and g(x) from 0 to z equal to each other, and then solving for z.
So, you need to solve the equation:
∫[0 to z] (1+sin(2x)) dx = ∫[0 to z] e^(x/2) dx
Once you solve this equation, the value of z you obtain will be the x-coordinate at which the shaded region R is divided into two equal areas.
For part C, your calculation is correct. The volume of the solid generated when R is revolved around the x-axis is approximately 4.267 cubic units.
For part D, the equation should actually be:
V = π∫[0 to 1.136] [(1+sin(2x))^2 - (e^(x/2))^2] dx + π∫[1.136 to ∞] [(1+sin(2x)) - π]^2 dx
This is because when revolving around the line x=π, the region R is divided into two parts: one from 0 to 1.136 (where g(x) is the outer function) and another from 1.136 to infinity (where f(x) is the outer function).
For part E, when revolving around the line y=π, the bounds of the integral will depend on the points of intersection between the graphs of f(x) and g(x) with the line y=π. You need to find those points of intersection and use them as the limits of integration in the volume integral.