Find the volume of the solid generated by revolving the region bounded by the graphs of the quations about the x-axis.
Y=x^2+4
Y=-x^2+2x+8
X=0
X=3
The region is lens-shaped, with tips at (-1,5) and (2,8). Unfortunately, our domain crosses over the right end of the enclosed region, ending at the strip from (3,7) to (13)
Let's say
y1 = -x^2+2x+8 and y2 = x^2+4
Using discs of thickness dx, we have
v = ∫π(R^2-r^2) dx
= ∫[0,2] π((y1)^2-(y2)^2) dx + ∫[2,3] π((y2)^2-(y1)^2) dx
Using shells of thickness dy, we have
v = ∫2πrh dy
where r=y and h is the horizontal distance between the curves x1 and x2, which gets a bit more involved.
followup. Using discs,
v = 272π/3 + 191π/3 = 463π/3
Using shells, it's a bit stranger. I made a typo, and the right boundary goes from y=5 to 13.
v
= ∫[4,8] 2πy√(y-4) dy
+ ∫[8,9] 2πy(1+√(9-y)-(1-√(9-y))) dy
+ ∫[5,8] 2πy(3-(1+√(9-y))) dy
+ ∫[8,13] 2πy(3-√(y-4)) dy
= 1024π/15 + 112π/5 + 94π/5 + 673π/15
= 463π/3
see
http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2%2B4,+y%3D-x%5E2%2B2x%2B8
To find the volume of the solid generated by revolving the region bounded by the graphs of the given equations about the x-axis, we can use the method of cylindrical shells.
First, let's graph the given equations and identify the region bounded by them:
1. Determine the x-coordinates where the curves intersect:
Set both equations equal to each other and solve for x:
x^2 + 4 = -x^2 + 2x + 8
Simplify the equation:
2x^2 - 2x - 4 = 0
Divide the equation by 2 to simplify:
x^2 - x - 2 = 0
Factor the quadratic equation:
(x - 2)(x + 1) = 0
Therefore, x = 2 and x = -1.
2. Plot the graph of each equation within the given x-range:
The equation y = x^2 + 4 is a parabola opening upwards with the vertex at (0, 4), passing through (0, 4) and (3, 13).
The equation y = -x^2 + 2x + 8 is a parabola opening downwards with the vertex at (1, 9), passing through (0, 8) and (3, 8).
Here is a graph of the given equations between x = 0 and x = 3:
(Note: The shaded region represents the bounded region.)
14 | ____
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12 | /
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10 |
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8 |
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0 1 2 3
3. Now, let's imagine a horizontal slice through the region along the x-axis. As we revolve this slice around the x-axis, it will create a cylindrical shell.
4. The height of each cylindrical shell is given by the difference in y-coordinates between the two curves at the corresponding x-value. Thus, the height of each shell is (upper curve - lower curve).
5. The radius of each cylindrical shell is simply the x-value at that height.
6. The differential volume of a cylindrical shell is calculated as follows:
dV = 2π * radius * height * dx
7. To find the total volume, we integrate the differential volume across the x-values of the bounded region:
V = ∫[a,b] 2π * radius * height * dx
In this case, the interval [a, b] is from x = -1 to x = 2.
Now we can set up the integral for the volume calculation:
V = ∫[-1,2] 2π * x * ((x^2 + 4) - (-x^2 + 2x + 8)) dx
Simplifying further:
V = ∫[-1,2] 2π * x * (2x^2 + 2x - 4) dx
Now, you can integrate this expression with respect to x from -1 to 2 to find the volume of the solid generated by revolving the region bounded by the graphs of the given equations about the x-axis.