Thread: 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?

Originally Posted by adam_leeds

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.
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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.

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.
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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.

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?

As Opalg said once you have the region it is up to you to decided the order of integration.



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.

Originally Posted by adam_leeds

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.

$\displaystyle \displaystyle \int_0^4\int_0^{4-x}3y\,dydx$ 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.

Cheers both of you.

Originally Posted by Opalg

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

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 $\displaystyle \displaystyle \int_0^4\int_0^{4-x}3y\,dydx.$