Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y = x + 2 and the parabola y = x^2 about the following lines:
a) The line x=2
b) The line x=-1
c) The x axis
d) The line y=26
I know what the formulas are, but I can't seem to apply them in this question. Please help me.
I'm trying to do this with just the x-axis but I keep getting the wrong answer to it. What am I supposed to do?
To apply the shell method to find the volume of the solid generated by revolving a region bounded by two curves, we need to integrate the circumference of the solid cross-sections.
We'll start by visualizing the region bounded by the line y = x + 2 and the parabola y = x^2. Let's find the intersection points of these two curves.
1) Find the intersection points of y = x + 2 and y = x^2:
x + 2 = x^2
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 and x = -1
So, the region is bounded by x = -1 and x = 2.
a) To find the volume of the solid generated by revolving this region about the line x = 2, we need to consider the horizontal shells.
1. Identify the axis of rotation: x = 2
2. Determine the radius of each shell: The radius is the distance between the axis of rotation and the curve at each particular x-value. In this case, the radius is x - 2.
3. Find the height of each shell: The height of each shell is the difference between the curves at a particular x-value. In this case, the height is (x + 2) - x^2.
4. Set up the integral: The integral represents the sum of the volumes of all the individual shells.
V = ∫ [2π(radius)(height)] dx, from x = -1 to x = 2
V = ∫ [2π(x - 2)((x + 2) - x^2)] dx, from x = -1 to x = 2
5. Evaluate the integral:
V = ∫ [2π(x - 2)(-x^2 + x + 2)] dx, from x = -1 to x = 2
V = ∫ [2π(-2x^3 +x^2 +2x -4)] dx, from x = -1 to x = 2
6. Integrate term by term:
V = [2π(-1/2x^4 + 1/3x^3 + x^2 - 4x)] from x = -1 to x = 2
V = [2π(1/3 + 4 - 4)] - [2π(-1/2 - 1/3 + 1 - 4)]
V = [2π(1/3)] - [2π(-5/6)]
V = [2π/3] + [5π/3]
V = 7π/3
Therefore, the volume of the solid generated by revolving the region bounded by y = x + 2 and y = x^2 about the line x = 2 is 7π/3.
You can follow a similar process to find the volumes for parts (b), (c), and (d) using the shell method.
To find the volume of the solid generated by revolving the region bounded by two curves about a given line using the shell method, you can follow these steps:
Step 1: Determine the limits of integration.
To find the limits of integration, we need to find the x-values at which the two curves intersect. Let's set the two curves equal to each other and solve for x:
x + 2 = x^2
Rearranging to set the equation to 0:
x^2 - x - 2 = 0
Factoring the quadratic equation:
(x - 2)(x + 1) = 0
This gives us two solutions: x = 2 and x = -1. Therefore, the limits of integration will be x = -1 and x = 2.
Step 2: Determine the radius.
The radius is the distance between the axis of revolution and the shell. In each case, the axis of revolution will be the given line.
a) The line x = 2:
The radius is the distance between x = 2 and the curve at a given x-value. In this case, the distance is x - 2.
b) The line x = -1:
Similarly, the radius is the distance between x = -1 and the curve at a given x-value. In this case, the distance is x + 1.
c) The x-axis:
In this case, since we are revolving about the x-axis, the radius is the y-value of the curve. The distance is simply y.
d) The line y = 26:
The radius is the distance between y = 26 and the curve at a given x-value. In this case, the distance is y - 26.
Step 3: Set up the integral.
The general formula for the shell method is:
V = 2π ∫ [radius * height] dx
a) The line x = 2:
The height is the difference between the upper and lower curves, which is (x + 2) - (x^2). Therefore, the integral becomes:
V_a = 2π ∫ [((x + 2) - (x^2)) * (x - 2)] dx, from x = -1 to x = 2
b) The line x = -1:
The height is still (x + 2) - (x^2), and the radius is x + 1. The integral becomes:
V_b = 2π ∫ [((x + 2) - (x^2)) * (x + 1)] dx, from x = -1 to x = 2
c) The x-axis:
Here, the radius is y = x^2 and the height becomes the difference between the x-axis and the curve, which is y. The integral becomes:
V_c = 2π ∫ [x^2 * x] dx, from x = -1 to x = 2
d) The line y = 26:
The radius is y - 26, and the height is (x + 2) - (x^2). The integral becomes:
V_d = 2π ∫ [((x + 2) - (x^2)) * (y - 26)] dx, from x = -1 to x = 2
Step 4: Evaluate the integrals.
Integrate each expression and substitute the limits of integration to calculate the volumes.
a) V_a: Integrate [(x + 2 - x^2) * (x - 2)] dx from -1 to 2.
b) V_b: Integrate [(x^2 - x - 2) * (x + 1)] dx from -1 to 2.
c) V_c: Integrate [(x^3)] dx from -1 to 2.
d) V_d: Integrate [(x + 2 - x^2) * (26 - x^2)] dx from -1 to 2.
Once you evaluate these integrals, you will find the volumes of the solids generated by revolving the region bounded by the given curves about the specified lines.
the graphs intersect at (-1,1) and (2,4)
Recall that the volume of a shell of thickness dx is
v = 2πrh dx
So, to rotate around the line x=2,
v = ∫[-1,2] 2πrh dx
where r = 2-x and h=(x+2)-x^2
v = ∫[-1,2] 2π(2-x)(x+2-x^2) dx = 27π/2
Just to verify, let's do it with discs (washers) as well. This one is bit more complicated, since the left boundary changes from the parabola to the line.
v = ∫π(R^2-r^2) dy
we have to break it up into two intervals.
#1: v = ∫[0,1] π(R^2-r^2) dy
where R=2+√y, r=2-√y
v = ∫[0,1] π((2+√y)^2-(2-√y)^2) dy = 16π/3
#2:
v = ∫[1,4] π(R^2-r^2) dy
where R=2-(y-2), r=2-√y
v = ∫[1,4] π((2-(y-2))^2-(2-√y)^2) dy = 49π/6
16π/3 + 49π/6 = 81π/6 = 27π/2
Looks like the shells were ok
Now do the others similarly.
If you type your integrals as I did, you can enter them in at wolframalpha.com and it will evaluate them, as well as show the indefinite integrals involved.