find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
x=2=2sqrty, x=0, y=9 about the y axis.
Is this the disk or shells method and how do I set this problem up?
first, clear up the meaning of
x=2=2sqrty
If you mean
x = 2+2√y, then since the curve intersects y=9 at (8,9) we have as usual, two ways to solve it. Using shells of thickness dx,
v = ∫[0,8] 2πrh dx
where r = x and h = 9-y = 9-(x/2 - 1)^2
v = 2π∫[0,8] x(9-(x/2-1)^2) dx = 1024/3 π
Of course, you can also use discs of thickness dy, but you have to allow for the hole in the middle for x in [0,2].
v = ∫[0,1] π(R^2-r^2) dy + ∫[1,9] πr^2 dy
where R = 2+2√y and r = 2-2√y
and where r = x = 2+2√y
I'll let you check my math and verify that they are equal.
To find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, we will use the disk method.
The equation x = 2 = 2sqrt(y) describes a semi-circle in the first quadrant. The curves x = 0 and y = 9 bound the region in the xy-plane.
To set up the problem, you can follow these steps:
1. Sketch the region in the xy-plane. You should see a semi-circle with radius 2, centered at (0,0), and bounded by the lines x = 0 and y = 9.
2. Determine the limits of integration. Since we are revolving the region around the y-axis, the limits of integration will be the y-values that bound the region. In this case, the region is bounded by y = 0 and y = 9.
3. Determine the radius of the infinitesimally thin disks. Since the region is being revolved around the y-axis, the radius of each disk will be the x-coordinate of the semi-circle at the corresponding y-value. In this case, the radius of the disk at y will be x = 2sqrt(y).
4. Set up the integral to calculate the volume using the disk method. The volume of a thin disk is given by the formula pi * (radius)^2 * (thickness). In this case, the radius is 2sqrt(y), and the thickness is dy (since we are integrating with respect to y). So the integral to find the volume is:
V = integral from y = 0 to y = 9 of (pi * (2sqrt(y))^2 * dy)
Simplifying and solving this integral will give you the volume of the solid.
To determine whether to use the disk or shell method for finding the volume of the solid obtained by rotating the region bounded by the given curves, we need to examine the shape of the region.
In this case, the region is bounded by the curves:
- x = 2
- x = 2√y
- x = 0 (y-axis)
- y = 9
Let's first sketch the curves to get a better understanding of their shape.
We can observe that the region is a semi-circle in the first quadrant, with the vertical line x = 2 bounding the right side and the x-axis (y = 0) bounding the bottom.
Now let's discuss the choice between the disk and shell method:
1. Disk method: This method involves slicing the solid into circular disks perpendicular to the axis of rotation (in this case, the y-axis). It is more suitable when the region is defined in terms of y.
2. Shell method: This method involves slicing the solid into cylindrical shells parallel to the axis of rotation. It is more suitable when the region is defined in terms of x.
In this problem, since the given region is defined in terms of x (x = 2√y), we will use the shell method.
Next, let's set up the problem using the shell method:
1. We need to determine the limits of integration. Since the region is bounded by the curves x = 2√y, x = 0, and y = 9, we need to find the values of y that define the region. From the equation x = 2√y, we can solve for y:
x = 2√y
x^2 = (2√y)^2
x^2 = 4y
y = (x^2)/4
The region is bounded by y = 0 and y = 9, so the limits of integration for y are 0 to 9.
2. Now, we will represent the volume element as a cylindrical shell with height Δy, radius of the shell at y = x, and thickness Δx.
- The height Δy corresponds to the change in y within the limits of integration (0 to 9).
- The radius of the shell at y = x is given by the curve x = 2√y.
- The thickness Δx will be infinitesimally small.
3. The volume of an individual shell is given by 2πrhΔx, where r is the radius of the shell (x = 2√y) and h is the height of the shell (Δy).
4. To find the total volume, we integrate the volume elements over the given range of y:
V = ∫(2πrh)dy
= ∫(2π(2√y)(y))dy
= ∫(4πy^(3/2))dy
= 4π∫(y^(3/2))dy
5. Evaluating the integral, we get:
V = 4π[(2/5)y^(5/2)] evaluated from 0 to 9
V = 4π[(2/5)(9^(5/2))] - 4π[(2/5)(0^(5/2))]
V = 4π[(2/5)(9^(5/2))]
V = (8/5)π(9^(5/2))
Therefore, the volume of the solid obtained by rotating the region about the y-axis is (8/5)π(9^(5/2)).