# Simple Double Integral

1. Nov 28, 2009

### Saladsamurai

1. The problem statement, all variables and given/known data

I don't know what is going on on my brain. I am at a sage in a problem where I need to evaluate the double integral:

$$\int\int_S(x+z)\,dS$$

where the surface is the is the portion of the plane x+y+x=1 that lies in the 1st octant.

3. The attempt at a solution

Forging ahead I arrive at:

$$\sqrt{3}\int\int_R(1-y)\,dx\,dy$$

I know that I am correct up until this point. The text verifies up till this step. So clearly my trouble lies in evaluating this simple double integral.

The text says the answer is 1/$\sqrt3$. I keep getting:

$$\sqrt{3}\int\int_R(1-y)\,dx\,dy$$

$$=\sqrt{3}\int_{x=0}^1\int_{y=0}^1(1-y)\,dx\,dy$$

$$=\sqrt{3}\int_{x=0}^1\left(\int_{y=0}^1(1-y)\,dy\right)dx$$

Before I go any further, is this last step ok? I am having trouble recalling whether it is ok or not to switch the order of integration from 'dxdy' to 'dydx' ?

I can't think of a reason why it wouldn't be......but it has been awhile. From here, I keep getting $\sqrt{3}/2$ which does not match the text's answer.

Thanks!

EDIT: I just tried it reversing the order which yields the same answer. I am convinced that my bounds are correct. Is the text just wrong?

Last edited: Nov 28, 2009
2. Nov 28, 2009

### Dunkle

Wouldn't your y bounds have to be from 0 to 1-x?

Edit: In the xy-plane, the "shadow" of the plane x+y+z = 1 will look like the triangle bounded by x=0, y=0 and x+y = 1.
Thus, if you say x is bounded by 0 and 1, then when you hold an arbitrary x constant, y is bounded by 0 and 1-x.

I worked out the integral with these new bounds and got the same answer as the text.

Last edited: Nov 28, 2009
3. Nov 28, 2009

### Saladsamurai

Hmmm. You are conjuring up distant memories here.... but I am still failing to recall why we do this.

Are the 'intercepts' of the plane with the x and y axes not fixed? Why do we need to write the y bounds in terms of x?

If I was near a text, I would try to find the answer, but unfortunately I am out at a coffee shop right now. Delicious.

Edit: I am examining you edit right now

4. Nov 28, 2009

### Saladsamurai

I am getting there, but I am still confused. I am still not understanding the answer to this "Are the 'intercepts' of the plane with the x and y axes not fixed?"

What I was doing was letting x and z equal 0 to find the y-bounds and letting y and z equal 0 to find the x-bounds. I know that that is not right now.... but I am not seeing why.

And why are the x-bounds from 0 to 1 correct? that is, why not choose the y-bounds to run from 0 to 1 and write the x-bounds in terms of y?

thanks,
Casey

5. Nov 28, 2009

### Dunkle

The intercepts are indeed fixed. It would also be perfectly fine to let y go from 0 to 1 and then let x go from 0 to 1-y if we switch the dydx to dxdy.

Maybe this will help: Draw out the region in the xy-plane. Now, we know the bounds for the outermost integral must be constants (or else we would end up with variables in our answer). So, let's choose x to go from 0 to 1. Now, to figure out the y bounds, we must fix an arbitrary x - say x is roughly 1/2. Now, draw a vertical line at x=1/2 and see what y is "cut off" by. You will notice that y is cut off by the line y=0 and the line y=1-x. So, we've found our bounds. The only time y goes from 0 to 1 is if x is 0. But if we chose x to be 0, we didn't pick a very arbitrary x ;)

6. Nov 28, 2009

### Saladsamurai

Oh.....I am so slow. The bounds that I gave are that of a rectangle :rofl:

It is getting clearer now. thanks for your patience Dunkle

7. Nov 28, 2009

### Saladsamurai

I am a horrible person. I thought it would be a piece of cake after getting the bounds right. I can't figure out where I am messing up the procedure:

$$=\sqrt{3}\int_{y=0}^{1-x}\left(\int_{x=0}^1(1-y)\,dx\right)\,dy$$

$$=\sqrt{3}\int_{y=0}^{1-x}\left[x - xy\right]_0^1\,dy$$

$$=\sqrt{3}\int_{y=0}^{1-x}(1-y)\,dy$$

$$=\sqrt{3}\left[y-\frac{y^2}{2}\right]_0^{1-x}$$

From here, I can see that I am going to be left with 'x' in the answer..... where am I fidging this now?

EDIT: I just found this on a website and also notice that Dunkle alluded to this point earlier: "In a double integral, the outer limits must be constant, but the inner limits can depend on the outer variable."

Can someone give a 'precise' reason as to why this is the case? I know, now, that the integral simply cannot be evaluated properly if the outer bounds are not constants....but what is the 'mathematical' reasoning for this?

Last edited: Nov 28, 2009
8. Nov 28, 2009

### Dunkle

Your integral should be either

$$\sqrt{3}\int_0^1 \int_0^{1-x} 1-y \ dydx \ \ \ \ or \ \ \ \ \sqrt{3}\int_0^1 \int_0^{1-y} 1-y \ dxdy$$

I think the second is easier to integrate, so let's do it!

$$\sqrt{3}\int_0^1 \int_0^{1-y} 1-y \ dxdy$$

$$= \sqrt{3}\int_0^1 \left[x-xy\right]_0^{1-y} \ dy$$

$$= \sqrt{3} \int_0^1 (1-y)-(1-y)y \ dy$$

$$= \sqrt{3} \int_0^1 y^{2}-2y+1 \ dy$$

$$= \sqrt{3} \left[\frac{1}{3}y^{3}-y^{2}+y\right]_0^1$$

$$= \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$

Edit:

Exactly! You could even call it a square

I don't have a "precise mathematical reason" for this, but you could think about it this way. Consider a region with area A in the xy-plane. If one were to integrate the function 1 over that region, the answer would be A (this follows from the definition of double integrals). If one did not have constant bounds on the outer integral, the answer would depend on either x or y. However, the area is always A and does not depend on x or y.

Last edited: Nov 28, 2009
9. Nov 28, 2009

### Saladsamurai

That's a great explanation. Thank you Dunkle. Now I feel like before I continue with this vector calculus text, I need to do a serious review of my multivariable calculus.

I hope PF is ready for an onslaught of questions

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