# Thread: Hoping I'm getting the hang of this: ∫∫ x^2 + 3y^2

1. ## Hoping I'm getting the hang of this: ∫∫ x^2 + 3y^2

z=x^2 +3y^2 ; x=0, y=1, y=x, z=0

I drew a diagram and figured out that y should be between 1 and x

From the diagram, it looks pretty clear to me that the lower bound of x should be 1. However, I am having difficulty figuring out what the upper bound should be. I would imagine that it should be the intersection between our function z=x^2+3y^2 and either y=x or y=1. However, plugging in z=0 would give me x^2=-3y^2 which seems really bad since I can't take the square root of a negative.

Thanks for your help guys ><

2. Originally Posted by crabchef
z=x^2 +3y^2 ; x=0, y=1, y=x, z=0

I drew a diagram and figured out that y should be between 1 and x

From the diagram, it looks pretty clear to me that the lower bound of x should be 1. However, I am having difficulty figuring out what the upper bound should be. I would imagine that it should be the intersection between our function z=x^2+3y^2 and either y=x or y=1. However, plugging in z=0 would give me x^2=-3y^2 which seems really bad since I can't take the square root of a negative.

Thanks for your help guys ><
The integral should be

$\displaystyle \int_0^{1}\int_x^{1}f(x,y)~dy~dx$

3. thanks for the answer...but could you please tell me how you arrived to it? I'm having trouble in how to determine what the bounds are

4. Originally Posted by crabchef
thanks for the answer...but could you please tell me how you arrived to it? I'm having trouble in how to determine what the bounds are
Think about the curves $\displaystyle y=1$ and $\displaystyle y=x$ except only consider to the right of the line x=0. Now these two curves intersect at x=1 so we can see that if we let $\displaystyle f(x)=x$ and $\displaystyle g(x)=1$ that

$\displaystyle \forall{x}\in(0,1)\quad{f(x)<g(x)}$

So there you go.

5. ah ok got it. The bounded area to the left is not x=1 but x=0. x=1, or the intersection between y=x and y=1 is the right hand bound.

thanks!