Let R be the region in the first quadrant bounded by the graphs of y=e^-x, y=1/2x+1, and x=2.
Then find the volume of the solid when R is revolved about each of the following lines... the x axis, y=-1, y=2, the y axis, x=-1, and x=3.
To find the volume of the solid when R is revolved about a given line, we can use the method of cylindrical shells.
1. Revolve around the x-axis:
- Identify the range of x-values for the region R. In this case, it is from x = 0 to x = 2.
- Select an arbitrary value, x, in the range of x-values and find corresponding y-values for the two curves y = e^(-x) and y = 1/2x + 1.
- The height of the cylindrical shell at x is the difference between these two y-values: h = (1/2x + 1) - e^(-x).
- The radius of the cylindrical shell is the x-value: r = x.
- The differential element of volume is given by dV = 2πrh dx = 2πx((1/2x + 1) - e^(-x)) dx.
- Integrate this expression over the range of x-values from 0 to 2: V = ∫[0 to 2] 2πx((1/2x + 1) - e^(-x)) dx.
2. Revolve around the line y = -1:
- In this case, we need to shift the region R upwards by 1 unit, so that when revolved, it is around the x-axis.
- The new curves will be y = e^(-x) + 1 and y = 1/2x + 1 + 1 = 1/2x + 2.
- Proceed with the same steps as in case 1 to find the volume.
3. Revolve around the line y = 2:
- In this case, we need to shift the region R downwards by 2 units, so that when revolved, it is around the x-axis.
- The new curves will be y = e^(-x) - 2 and y = 1/2x + 1 - 2 = 1/2x - 1.
- Proceed with the same steps as in case 1 to find the volume.
4. Revolve around the y-axis:
- Identify the range of y-values for the region R. In this case, it is from y = 0 to y = 1. (Note that the curves are inverted since we're rotating around the y-axis).
- Select an arbitrary value, y, in the range of y-values and find corresponding x-values for the two curves y = e^(-x) and y = 1/2x + 1.
- The height of the cylindrical shell at y is the difference between these two x-values: h = (1/2x + 1) - e^(-x).
- The radius of the cylindrical shell is the y-value: r = y.
- The differential element of volume is given by dV = 2πrh dy = 2πy((1/2x + 1) - e^(-x)) dy.
- Integrate this expression over the range of y-values from 0 to 1: V = ∫[0 to 1] 2πy((1/2x + 1) - e^(-x)) dy.
5. Revolve around the line x = -1:
- In this case, we need to shift the region R to the right by 1 unit, so that when revolved, it is around the y-axis.
- The new curves will be x = -1 + e^(-y) and x = -1 + 1/2y + 1 = -1 + 1/2y + 2 = 1/2y + 1.
- Proceed with the same steps as in case 4 to find the volume.
6. Revolve around the line x = 3:
- In this case, we need to shift the region R to the left by 3 units, so that when revolved, it is around the y-axis.
- The new curves will be x = 3 + e^(-y) and x = 3 + 1/2y + 1 = 3 + 1/2y + 2 = 1/2y + 5.
- Proceed with the same steps as in case 4 to find the volume.
By following these steps for each case, you can calculate the volume of the solid when R is revolved about each given line.
To find the volume of the solid when the region R is revolved about a certain line, we can use the disk or washer method. Let's consider each line one by one:
1. Revolving about the x-axis:
In this case, we can use the disk method. The volume of each infinitesimally thin disk is given by πy^2dx, where y is the distance from the line of revolution (in this case, the x-axis). We need to integrate this expression over the appropriate range of x.
To find the range of integration, we need to determine the x-values at which the curves intersect.
For the curves y=e^-x and y=1/2x+1, we can set them equal to each other:
e^-x = 1/2x + 1
We can solve this equation numerically using a graphing calculator or software to find the x-value of intersection, which turns out to be approximately x = 0.3517.
Therefore, the integral for the volume when revolving about the x-axis is:
V_x-axis = ∫[0 to 2] π(e^-x)^2dx
2. Revolving about the line y = -1:
In this case, the line of revolution is below the region R, so we can use the washer method. The volume of each washer is given by π(R_outer^2 - R_inner^2)dx, where R_outer and R_inner represent the radii of the outer and inner edges of the washer, respectively.
To find the radii, we need to determine the y-values at the x-values of intersection of the curves. We again solve the equation e^-x = 1/2x + 1 to find the x-value, which is approximately 0.3517 as mentioned before.
The integral for the volume when revolving about y = -1 is:
V_y=-1 = ∫[0 to 2] π((1/2x+1)^2 - (e^-x)^2)dx
3. Revolving about the line y = 2:
In this case, the line of revolution is above the region R, so we again use the washer method. The radii of the washers are calculated similarly as in the previous case. The x-value of intersection between y = e^-x and y = 1/2x + 1 is approximately 0.3517.
The integral for the volume when revolving about y = 2 is:
V_y=2 = ∫[0 to 2] π((2 - 1/2x - 1)^2 - (e^-x)^2)dx
4. Revolving about the y-axis:
When revolving about the y-axis, we use the disk method. The volume of each disk is given by πx^2dy, where x is the distance from the line of revolution (in this case, the y-axis). We need to integrate this expression over the y-range of the region.
To find the range of integration, we need to determine the y-values at the x-values of intersection of the curves. We solve the equation e^-x = 1/2x + 1 to find the x-value, which is approximately 0.3517.
The integral for the volume when revolving about the y-axis is:
V_y-axis = ∫[0 to e^-x] πx^2dy
5. Revolving about the line x = -1:
In this case, the line of revolution is to the left of the region R, so we once again use the washer method. The radii of the washers can be determined by finding the x-values at the y-values of intersection. Solving e^-x = 1/2x + 1, we find the intersection x-value approximately as 0.3517.
The integral for the volume when revolving about x = -1 is:
V_x=-1 = ∫[0 to e^-x] π((1/2x+1)^2 - (x+1)^2)dy
6. Revolving about the line x = 3:
In this case, the line of revolution is to the right of the region R, so we again use the washer method. Similar to previous cases, the radii of the washers are determined by finding x-values at y-values of intersection. The x-value of intersection between y = e^-x and y = 1/2x + 1 is approximately 0.3517.
The integral for the volume when revolving about x = 3 is:
V_x=3 = ∫[0 to e^-x] π((x+3)^2 - (1/2x+1)^2)dy
By evaluating these integrals numerically or using appropriate software, you can find the volumes of the solids when R is revolved about each of the given lines.