# Double Integration, finding Area

1. Mar 12, 2008

### dalarev

[SOLVED] Double Integration, finding Area

1. The problem statement, all variables and given/known data

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

2. Relevant 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.

3. 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.

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Last edited: Mar 12, 2008
2. Mar 12, 2008

### HallsofIvy

Staff Emeritus
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
$$\int_{x= -2}^4\int_{y= x^2}^{2x+ 8} x dydx$$

3. Mar 12, 2008

### dalarev

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