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July 31st, 2017, 06:45 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 yaxis 
July 31st, 2017, 07:04 AM  #2 
Senior Member Joined: Oct 2009 Posts: 232 Thanks: 82 
You're going to have to be more detailed than this!

July 31st, 2017, 07:40 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,700 Thanks: 1358  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 yaxis using cylindrical shells ... $\displaystyle 2\pi \int_0^5 x[(322x)(2+4x)] \, dx$ using disks about the yaxis will require two integral expressions ... $\displaystyle \pi \int_2^{22} \left(\dfrac{y2}{4}\right)^2 \, dy + \pi \int_{22}^{32} \left(\dfrac{32y}{2}\right)^2 \, dy$ 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. 
July 31st, 2017, 03:32 PM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,944 Thanks: 797 
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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