# Thread: Bounded Double Integral Problem.

1. ## Bounded Double Integral Problem.

So I'm trying to do a problem for a hw, and I can't seem to get the proper answer

The problem is giving a triangle, whose side are y from 0 to 2 and x from 0 to 4.

It wants me to integrate the function f(x,y) = x^2y within this boundary

So I chose for my constant of this function to be y is less than or equal to 2x and greater than or equal to 0. This is confirmed by an example in the book using an identical triangle

So i integrated the function with respect to x using these limit and go 8x^4/3
I then integrated between 0 and 2 with respect to y and go 256/15

this isnt right, and I cant seem to get a different answer than this

any help?

2. Originally Posted by TheUnfocusedOne
So I'm trying to do a problem for a hw, and I can't seem to get the proper answer

The problem is giving a triangle, whose side are y from 0 to 2 and x from 0 to 4.

It wants me to integrate the function f(x,y) = x^2y within this boundary

So I chose for my constant of this function to be y is less than or equal to 2x and greater than or equal to 0. This is confirmed by an example in the book using an identical triangle

So i integrated the function with respect to x using these limit and go 8x^4/3
I then integrated between 0 and 2 with respect to y and go 256/15

this isnt right, and I cant seem to get a different answer than this

any help?

According to what you describe, it seems to me that the triangle's vertices are (0,0), (0,2), (4,0) ==> the straight line joining these last two vertices is $\displaystyle y=-\frac{x}{2}+2$, so you can put the following limits:

$\displaystyle 0\leq x\leq 4\,,\,\,-\frac{x}{2}+2\leq y\leq 0$, and thus you want:

$\displaystyle \int\limits_0^4\int\limits_0^{-\frac{x}{2}+2}\!\!x^2y\,dydx=\frac{1}{2}\,\int\lim its_0^4x^2\left(\left(-\frac{x}{2}+2\right)^2\right)dx=\frac{64}{15}$ (I think)

I don't understand why you put $\displaystyle 0\leq y\leq 2x$ when x goes from 0 to 4...

Tonio

3. Im not sure why you chose to do -x/2 + 2 when you could of just done x/2, serves the same purpose in my mine since when x = 4, y = 2 and when x = 0, y = 0.

The boundaries I set were the same as the book had set. Im not trying to be rude but I kinda wanna follow the book seeing as though thats where the problems are from.

4. Originally Posted by TheUnfocusedOne
Im not sure why you chose to do -x/2 + 2 when you could of just done x/2, serves the same purpose in my mine since when x = 4, y = 2 and when x = 0, y = 0.

The boundaries I set were the same as the book had set. Im not trying to be rude but I kinda wanna follow the book seeing as though thats where the problems are from.

It's not that x/2 "works": it's just incorrect: the boundaries of the problem YOU gave form a triangle with vertices at (0,0), (0,2), (4,0), and these points are joined by the lines x = 0, 0 <= y <=2 (joining (0,0), (0,2)), the line y = 0, 0 <= x <= 4 (joining (0,0), (4,0)), and the line y = -x/2 + 2, joining (0,2) and (4,0).

The line y = x/2 passes thru the origin and the point (4,2), and this last one is not in your triangle.

What book do you use, anyway?

Tonio

5. Multivariable Calculus by Jon Rogawski

Its only the last section of a much large book that brings one through calcs 1-3

6. attaching a photo of the triangle

7. Originally Posted by TheUnfocusedOne
attaching a photo of the triangle

Ooh, I see! Your first post misled me since it indeed is y between zero and 2 BUT x is between 2y and 2.
Anyway, I did it in both forms and I get the same result as you did: 256/15, so I can't tell you why the book says otherwise.

Tonio