Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the line y=3.
y=√x-1 ,y=0 ,x=5
OK - I'll do one more, then it's your turn
using shells of thickness dy, we have
v = ∫[0,2] 2πrh dy
where r = 3-y and h = 5-x = 5 - (y^2+1)
v = ∫[0,2] 2π(3-y)(5-(y^2+1)) dy = 24π
you can check your answer using discs of thickness dx:
v = ∫[1,5] π(R^2-r^2) dx
where R = 3 and r = 3-y = 3-√(x-1)
v = ∫[1,5] π(3^2-(3-√(x-1))^2) dx = 24π
Question, where did the 3-y come from. Also, I will do the beginning of the other problem I requested help for, but I am trying to understand how you got 3-y. I understand how you got your height, but not the 3-y.
the axis of rotation is y=3, but the graph is some distance away from it. The radius of the shells is the distance from the curve to the axis of rotation.
Did you not sketch the curves in question?
I did, but I wasn't sure if I was supposed to subtract the y from the axis of rotation. I was in doubt of myself on that one. I understand.
To find the volume generated by rotating the region bounded by the given curves about the line y=3, we can use the method of cylindrical shells. This involves integrating the circumference of each cylindrical shell multiplied by its height.
1. First, let's draw a diagram to visualize the region bounded by the curves y=√x-1, y=0, and x=5.
- On the x-y plane, we have the x-axis, the y-axis, and the line y=3.
- The curve y=√x-1 represents a parabola opening upwards, passing through the point (1, 0).
- The curve y=0 represents the x-axis.
- The line x=5 represents a vertical line passing through the point (5, 0).
The region bounded by these curves can be visualized as the area under the curve y=√x-1, starting from x=1 to x=5, and above the x-axis.
2. Now, let's consider an infinitesimally small vertical strip (or cylindrical shell) within this region.
- Let's take an infinitesimally small section dx along the x-axis.
- The height of the cylindrical shell will be the difference between the y-coordinate of the curve y=√x-1 and the line y=3.
- The radius of the cylindrical shell will be the x-coordinate.
3. To find the volume of the cylindrical shell, we need to calculate its circumference and multiply it by its height.
- The circumference of the cylindrical shell is 2π times the radius, which is 2πx.
- The height of the cylindrical shell is (√x-1) - 3, or √x-4.
4. To find the volume of the entire region, we need to integrate the volume of each cylindrical shell over the bounds from x=1 to x=5.
- The integral expression for the volume V is:
V = ∫(2πx)(√x-4) dx, evaluated from x=1 to x=5.
5. Evaluate the integral to find the volume.
- Integrate the expression (2πx)(√x-4) with respect to x over the bounds 1 to 5.
- V = ∫(2πx)(√x-4) dx = 2π ∫(x√x - 4x) dx, evaluated from x=1 to x=5.
- Using the power rule for integration, simplify the integral expression and evaluate it over the bounds.
- V = 2π [(2/5)x^(5/2) - 2x^2] evaluated from x=1 to x=5.
6. Calculate the volume V.
- Plug in the upper and lower limits into the expression and subtract to find the volume.
- V = 2π [(2/5)(5^(5/2)) - 2(5^2)] - 2π [(2/5)(1^(5/2)) - 2(1^2)].
- Simplify the expression and calculate the value of V.
Thus, by following these steps and evaluating the given integral, you can find the volume generated by rotating the region bounded by the curves about the line y=3 using the method of cylindrical shells.