Thread: Double Integral and trying to find the region 'D'

I have the double integral of...
∫∫y^3 dA

and the region 'D' are the points:
(0,2),(1,1),(3,2)

I graphed it, it makes a triangle but how in the world can I set up my integral like this? I really don't see an upper and lower equation, but I do see the ends of the region I can take.

If I fix 'x' I can go from 0 to 3
If I fix 'y' I can go from 1 to 2

but now what do I find? or how do I find it for that matter...

Originally Posted by RET80

I have the double integral of...
∫∫y^3 dA

and the region 'D' are the points:
(0,2),(1,1),(3,2)

I graphed it, it makes a triangle but how in the world can I set up my integral like this? I really don't see an upper and lower equation, but I do see the ends of the region I can take.

If I fix 'x' I can go from 0 to 3
If I fix 'y' I can go from 1 to 2

but now what do I find? or how do I find it for that matter...

Always go back to the general form of the double integral



Now draw the triangle. We clearly see that the top function here is the line . What about a bottem function? Well we can see there are
2 parts here, one from x=0-->x=1 and one from x=1-->x=3

Let us note that we can model sides of a triangle via the line y=mx+b

by doing this we get

for the region

for the region

Now we can integrate our original function from these bounds.

So we get



This should render the desired result

Originally Posted by AllanCuz

Always go back to the general form of the double integral



Now draw the triangle. We clearly see that the top function here is the line . What about a bottem function? Well we can see there are
2 parts here, one from x=0-->x=1 and one from x=1-->x=3

Let us note that we can model sides of a triangle via the line y=mx+b

by doing this we get

for the region

for the region

Now we can integrate our original function from these bounds.

So we get



This should render the desired result

thanks, but why are the integrals from the y=mx+b equations to '2' ?

Originally Posted by RET80

thanks, but why are the integrals from the y=mx+b equations to '2' ?

We need both a floor and top function. By drawing the triangle we can see that the top function is y=2. Naturally when we follow the general double integral we are looking for a function in terms of x. But in this case which still has an "x" function if you want to think about it like that, but it has a co-efficient of zero. It's also of importance to note we are bounded by this region, by the lines (as given by y=mx+b) and y=2. This is what we want! Thus, our top is y=2.

Originally Posted by RET80

thanks, but why are the integrals from the y=mx+b equations to '2' ?

Two of the points, (0, 2) and (3, 2) have y-coordinate 2. One of the things you should have learned long ago is that two points determine a line and a line can be written as y= mx+ b. Taking x= 0 and y= 2 gives 2= b. Taking x= 3 and y= 3 gives 2= 3m+ b. Subtracting the first equation from the second gives 3m= 0 so m= 0 and b= 2. The equation of the top line is "y= 2".