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 July 31st, 2017, 05:45 AM #1 Newbie   Joined: Jul 2017 From: Earth Posts: 6 Thanks: 1 How do I find the limit of integration ? y = 32 - 2x y = 2 + 4x x = 0 about y-axis
 July 31st, 2017, 06:04 AM #2 Senior Member   Joined: Oct 2009 Posts: 428 Thanks: 144 You're going to have to be more detailed than this!
July 31st, 2017, 06:40 AM   #3
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Quote:
 Originally Posted by JoKo 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 06:49 AM.

 July 31st, 2017, 02:32 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,236 Thanks: 884 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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