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July 31st, 2017, 06:45 AM   #1
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How do I find the limit of integration ?

y = 32 - 2x
y = 2 + 4x
x = 0
about y-axis
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July 31st, 2017, 07:04 AM   #2
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You're going to have to be more detailed than this!
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July 31st, 2017, 07:40 AM   #3
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Quote:
Originally Posted by JoKo View Post
y = 32 - 2x
y = 2 + 4x
x = 0
about y-axis
Looks like a volume of revolution problem; did you make a sketch of the region?

the first two lines intersect at $x = 5$

rotating the region within the three lines about the y-axis using cylindrical shells ...

$\displaystyle 2\pi \int_0^5 x[(32-2x)-(2+4x)] \, dx$


using disks about the y-axis will require two integral expressions ...

$\displaystyle \pi \int_2^{22} \left(\dfrac{y-2}{4}\right)^2 \, dy + \pi \int_{22}^{32} \left(\dfrac{32-y}{2}\right)^2 \, dy$


you may check both by using the volume formula for the two cones formed with the common base, $r=5$ ... upper height $h=10$ and lower height $h = 20$

Last edited by skeeter; July 31st, 2017 at 07:49 AM.
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July 31st, 2017, 03:32 PM   #4
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Or it could be just an "integrate over this region" problem. This region is bounded by the three given lines, x= 0, y= 32- 2x, y= 2+ 4x. The lines x= 0 and y= 32- 2x cross at (0, 32), the two lines x= 0 and y= 2+ 4x cross at (0, 2), and, as Skeeter said, the lines y= 32- 2x= 2+ 4x cross when y= 32- 2x= 2+ 4x, 6x= 30, so x= 5, y= 22. That is, the region is the triangle with vertices (0, 2), (0, 32), and (5, 22).

One way to integrate is to take x going from 0 to 5, taking y going, for each x, from y= 2+ 4x to y= 32- 2x. To integrate f(x,y) over that region integrate $\displaystyle \int_0^5\int_{2+ 4x}^{32- 2x} f(x, y) dydx$
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