Which of the following integrals will find the volume of the solid that is formed when the region bounded by the graphs of y = ex, x = 1, and y = 1 is revolved around the line y = -2
I know it is
pi (integral from zero to 1)then I get confused
To find the volume of the solid that is formed when the region bounded by the graphs of y = ex, x = 1, and y = 1 is revolved around the line y = -2, we need to set up an appropriate integral.
The region bounded by the given graphs is a triangle, with a base along the x-axis from x = 0 to x = 1 and sides given by y = ex and y = 1.
To find the volume using the disc method, we need to consider the cross-sections of the solid. These cross-sections are disks with a radius determined by the distance between the curve and the axis of rotation (y = -2), and a thickness determined by the change in x.
Now, let's consider the disc method for integration:
1. Calculate the radius of each disc. The distance from the curve y = ex to the axis of rotation y = -2 is y + 2. Since the curve is given by y = ex, the radius of each disc is (ex + 2).
2. Calculate the area of each disc. The formula for the area of a disc is A = πr^2, where r is the radius. In this case, the area of each disc is π(ex + 2)^2.
3. Set up the integral for the volume. The volume can be calculated by integrating the area of each disc over the region. Since the base of the region is along the x-axis, the integral will be with respect to x. The limits of integration will be from x = 0 to x = 1.
After these steps, we can set up the integral:
V = ∫[0,1] π(ex + 2)^2 dx
Therefore, the integral to find the volume of the solid is ∫[0,1] π(ex + 2)^2 dx.
To find the volume of the solid formed when the region bounded by the graphs of y = ex, x = 1, and y = 1 is revolved around the line y = -2, we can use the disk method or the shell method.
The disk method is appropriate when the slices of the solid are perpendicular to the axis of revolution. The shell method is appropriate when the slices of the solid are parallel to the axis of revolution.
Let's go through both methods and determine the correct integral.
Disk Method:
1. Draw a vertical line passing through the region bounded by the graphs.
2. For each x-value from x = 0 to x = 1, calculate the radius of the disk by finding the vertical distance from y = ex to y = 1.
- The radius of the disk is given by r(x) = 1 - ex.
3. The area of each disk is given by A(x) = π * [r(x)]^2.
4. Integrate the area function A(x) from x = 0 to x = 1 to find the volume:
- V = ∫[0 to 1] A(x) dx = ∫[0 to 1] π * [1 - ex]^2 dx
Shell Method:
1. Draw a horizontal line passing through the region bounded by the graphs.
2. For each y-value from y = -2 to y = 1, calculate the radius of each shell by finding the horizontal distance from x = 0 to x = [ln(1 - y)].
- The radius of each shell is given by r(y) = ln(1 - y).
3. The height of each shell is given by h(y) = [1 - ex] - (-2).
4. The volume of each shell is given by V(y) = 2π * r(y) * h(y).
5. Integrate the volume function V(y) from y = -2 to y = 1 to find the total volume:
- V = ∫[-2 to 1] V(y) dy = ∫[-2 to 1] 2π * [ln(1 - y)] * [1 - ex] dy
Comparing the given options with the derived integrals, the integral that will find the volume of the solid when the region is revolved around y = -2 is:
- V = ∫[-2 to 1] 2π * [ln(1 - y)] * [1 - ex] dy
I assume you have drawn the region. It's a triangular shape with vertices at (0,1), (1,1) and (1,e).
So, using discs (washers) of thickness dx,
v = ∫[0,1] π(R^2-r^2) dx
where R=y+2 and r=3
v = ∫[0,1] π((e^x+2)^2-3^2) dx
= π/2 (e^2+8e-19)
Or, using shells of thickness dy,
v = ∫[1,e] 2πrh dy
where r=y+2 and h=1-x
v = ∫[1,e] 2π(y+2)(1-ln y) dy
= π/2 (e^2+8e-19)