Volume of rotating solid

J

jwpaine

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Problem: Find the volume of the solid obtained by rotating the region enclosed by the graphs about the given axis.
y = 16 - x, y = 3x + 12, x = 0, about the y-axis

MY work:

the graphs of y = 3x + 12 and y = 16 - x meet at the point (1, 15)
As I am using the disk method for this one and not the shell method, I am breaking the problem up into two separate integrals for computing disks of height dy from y = 12 up to y = 15, and then computing disks of height dy from y = 15 to y = 16

Let the area enclosed by y = 3x + 12, y = 15, and x = 0 be A1
Let the area enclosed by y = 16 - x, y = 15 and x = 0 be A2

For A1, the radius of the disks will be (x_right - x_left) y = 3x + 12 - 0, in terms of y: (y - 12) / 3 = x
For A2, the radius of the disks will be (x_right - x_left y = 16 - y - 0, in terms of y: (16 - y) = x

A1_dv = (Pi)(r)^2 * dy = (Pi)( (y - 12) / 3 )^2 * dy
A2_dv = (pi)(r)^2 * dy = (Pi)( 16 - y )^2 * dy

Total area = \(\displaystyle \pi \int_{12}^{15} (\frac{y - 12}{3})^2 \cdot dy \,\,\,+\,\,\, \pi \int_{15}^{16} (y^2 - 32y + 256) \cdot dy\)

What in the world have I done wrong?
Thanks!

John
 
Hey Galactus, check your shaded region. It should be the little triangle bounded by the two linear functions and x = 0, not y = 0... and rotation around the y-axis.

Do I have it right than?

Thanks so much!
 
jwpaine said:
Problem: Find the volume of the solid obtained by rotating the region enclosed by the graphs about the given axis.
y = 16 - x, y = 3x + 12, x = 0, also y = 0 ?about the y-axis

MY work:

the graphs of y = 3x + 12 and y = 16 - x meet at the point (1, 15)
As I am using the disk method for this one and not the shell method, I am breaking the problem up into two separate integrals for computing disks of height dy from y = 12 up to y = 15, and then computing disks of height dy from y = 15 to y = 16 I would suggest using disks from y = 0 to y = 12, then washers from y = 12 to y = 15

Let the area enclosed by y = 3x + 12, y = 15, and x = 0 be A1 This region is the hollowed out part of the solid. If you want to integrate this region, you would subtract it from a solid generated from integrating disks from y = 0 to y = 15 (because what I think you're calling A1 is the "empty" space in the concave part of the solid.)
Let the area enclosed by y = 16 - x, y = 15 and x = 0 be A2

For A1, the radius of the disks will be (x_left - x_right) y = 3x + 12 - 0, in terms of y: (y - 12) / 3 = x
For A2, the radius of the disks will be (x_left - x_right) y = 16 - y - 0, in terms of y: (16 - y) = x

A1_dv = (Pi)(r)^2 * dy = (Pi)( (y - 12) / 3 )^2 * dy
A2_dv = (pi)(r)^2 * dy = (Pi)( 16 - y )^2 * dy

Total area = \(\displaystyle \pi \int_{12}^{15} (\frac{y - 12}{3})^2 \cdot dy \,\,\,+\,\,\, \pi \int_{15}^{16} (y^2 - 32y + 256) \cdot dy\)
Click to expand...
 
area bounded by x = 0.... rotation about the y-axis. There is no hole. y = 0 has nothing to do with it.

Just as the problem described: Find the volume of the solid obtained by rotating the region enclosed by the graphs about the given axis.
y = 16 - x, y = 3x + 12, x = 0, about the y-axis
 
I'm sorry, JW. I misconstrued the problem. I see now. Let me get back.
 
After looking at what you have, that's how I done it. Did you get \(\displaystyle \frac{4{\pi}}{3}\)?.
 
Did you setup the integrals just like I had them? I might have made an error in my arithmetic.

Double checking.......

First int \(\displaystyle \pi \int_{12}^{15} (\frac{y-12}{3})^2 dy = \pi\)

Second int \(\displaystyle \pi \int_{15}^{16} (16-y)^2 dy = \frac{\pi}{3}\)

Well what do you know.........

\(\displaystyle \pi + \frac{\pi}{3} = \frac{4\pi}{3}\)

The back of my book confirms this.....

Thanks a lot, and sorry for wasting your time :oops: :D

John
 
You didn't waste my time. I may have wasted yours by posting the wrong region the first time.
Yes, I set it up the way you did. I think you just had an arithmetic error, as you said.
 
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