# Surface area double integrals

• Feb 24th 2011, 11:07 AM
Surface 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 PM
Opalg
Quote:

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?

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.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
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 AM
Quote:

Originally Posted by Opalg
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.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
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.

Thanks i have got this, but am a bit confused with the order of integration is the first integration i do (the inner one im guessing) dx or dy?
• Feb 25th 2011, 10:22 AM
TheEmptySet
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

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

If you integrate with respect to y first you will get

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

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 AM
Opalg
Quote:

Thanks i have got this, but am a bit confused with the order of integration is the first integration i do (the inner one im guessing) dx or dy?

That's right, you start from the inner integral and work outwards.

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

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 AM
$\displaystyle \int_0^4\int_0^{4-x}3y\,dydx$ means $\displaystyle \int_0^4\biggl(\int_0^{4-x}3y\,dy\biggr)dx$.
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 $\displaystyle \int_0^4\int_0^{4-x}3\,dydx$, not $\displaystyle \int_0^4\int_0^{4-x}3y\,dydx.$