f(x,y) = 3y located above the region bounded by the x-axes, y-axes and the line y = 4-x

How do i work out the limits for my integrals?

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- Feb 24th 2011, 11:07 AMadam_leedsSurface area double integrals
f(x,y) = 3y located above the region bounded by the x-axes, y-axes and the line y = 4-x

How do i work out the limits for my integrals? - Feb 24th 2011, 12:57 PMOpalg
The essential thing here is to DRAW A DIAGRAM of the region. It is bounded by the x- and y-axes and the line y = 4 – x.

__Draw that diagram before reading any further.__

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Done that? You should have drawn a triangular region. Now imagine that region sliced by vertical lines, which each correspond to a fixed value of x. For each value of x between x=0 and x=4, the line will pass through the region (if x<0 or x>4 then the line will miss the region altogether). The section of line within the region goes from y=0 up to y=4–x. So the (inner) y integral should go from 0 to 4–x and the (outer) x integral will go from 0 to 4. - Feb 25th 2011, 10:06 AMadam_leeds
- Feb 25th 2011, 10:22 AMTheEmptySet
As Opalg said once you have the region it is up to you to decided the order of integration.Attachment 20938

http://www.mathhelpforum.com/math-he...2&d=1298661341

If you integrate with respect to x first you will get

$\displaystyle \displaystyle \int _{0}^{4}\int_{0}^{4-y}f(x,y)dxdy$

If you integrate with respect to y first you will get

$\displaystyle \displaystyle \int _{0}^{4}\int_{0}^{4-x}f(x,y)dydx$

What this comes down to is do you want your first set of rectangles to be horizontal as opposed to vertical. - Feb 25th 2011, 10:23 AMOpalg
That's right, you start from the inner integral and work outwards.

$\displaystyle \displaystyle \int_0^4\int_0^{4-x}3y\,dydx$ means $\displaystyle \displaystyle \int_0^4\biggl(\int_0^{4-x}3y\,dy\biggr)dx$.

But as TheEmptySet says, you can do the integral in the reverse order if you carve up the region of integration using horizontal sections rather than vertical. - Feb 25th 2011, 10:28 AMadam_leeds
Cheers both of you.

- Feb 25th 2011, 11:31 AMOpalg
**Correction**(thanks to Krizalid for quietly pointing this out to me):

I was concentrating so much on the region of integration that I didn't read the rest of the question carefully. The integral that I gave is for the*volume*under the surface z = 3y. But the question asked for the*surface area*. That should be given by the integral $\displaystyle \displaystyle \int_0^4\int_0^{4-x}3\,dydx$, not $\displaystyle \displaystyle \int_0^4\int_0^{4-x}3y\,dydx.$