Double Integration, finding Area

[dalarev]

[SOLVED] Double Integration, finding Area

Homework Statement

Find: \int\int_{A} xdxdy , where A is the area between y=x^2 and y=2x+8

Homework Equations

The points of intersection of the two functions is at x=-2 and at x=4. Attached is a plot with the area asked to find.

The Attempt at a Solution

I'm seeing a problem with the x limits of integration changing at x=-2, one of their intersections. I am pretty sure this can be done by summing the 2 areas separately, but the problem asks to solve for the double integral.

 

[HallsofIvy]

IF you were to integrate \int \int x dx dy, yes, you would have to do it as two separate integrals: as y goes from 0 to 4, x must range between the two sides of the parabola, x= -\sqrt{y} and x= \sqrt{y}. For y between 4 and 16, x ranges from the straight line on the left to the parabola on the right: x= (1/2)y- 4 . The integral is given by
\int_{y=0}^4\int_{x= -\sqrt{y}}^{\sqrt{y}}x dxdy+ \int_{y= 4}^{16}\int_{x= y/2- 4}^{\sqrt{y}} xdxdy

Are you required to do it in that order? The other order, dydx, would seem to me simpler and more natural. In this case, x must range from -2 to 4 and, for every x, y ranges from x^2 to 2x+ 8. In that order, the integral is
\int_{x= -2}^4\int_{y= x^2}^{2x+ 8} x dydx

 

[dalarev]

Are you required to do it in that order? The other order, dydx, would seem to me simpler and more natural. In this case, x must range from -2 to 4 and, for every x, y ranges from x^2 to 2x+ 8. In that order, the integral is
\int_{x= -2}^4\int_{y= x^2}^{2x+ 8} x dydx

Yeah, that's what I ended up doing. I often get too caught up in small details that prevent me from even starting the problem. This was pretty straightforward once I reversed the order of integration. Once again, thanks for your thorough help. /SOLVED