Double Integral and trying to find the region 'D'

• Apr 23rd 2010, 02:51 PM
RET80
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...
• Apr 23rd 2010, 03:06 PM
AllanCuz
Quote:

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

$\int_a^b dx \int_{g_1(x)}^{g_2(x)} dy$

Now draw the triangle. We clearly see that the top function here is the line $y = 2$. 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

$y_1 = -1x + 2$ for the region $0 \le x \le 1$

$y_2 = \frac{1x}{2} + \frac{1}{2}$ for the region $1 \le x \le 3$

Now we can integrate our original function from these bounds.

So we get

$\iint y^3 dA = \int_0^1 dx \int_{-x + 2}^2 y^3 dy + \int_1^3 dx \int_{ \frac{1x}{2} + \frac{1}{2} }^2 y^3 dy$

This should render the desired result
• Apr 24th 2010, 04:14 PM
RET80
Quote:

Originally Posted by AllanCuz
Always go back to the general form of the double integral

$\int_a^b dx \int_{g_1(x)}^{g_2(x)} dy$

Now draw the triangle. We clearly see that the top function here is the line $y = 2$. 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

$y_1 = -1x + 2$ for the region $0 \le x \le 1$

$y_2 = \frac{1x}{2} + \frac{1}{2}$ for the region $1 \le x \le 3$

Now we can integrate our original function from these bounds.

So we get

$\iint y^3 dA = \int_0^1 dx \int_{-x + 2}^2 y^3 dy + \int_1^3 dx \int_{ \frac{1x}{2} + \frac{1}{2} }^2 y^3 dy$

This should render the desired result

thanks, but why are the integrals from the y=mx+b equations to '2' ?
• Apr 24th 2010, 05:50 PM
AllanCuz
Quote:

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 $y=2+0x$ 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.
• Apr 25th 2010, 03:37 AM
HallsofIvy
Quote:

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".