# Double Integral

1. Oct 1, 2006

### Saketh

I'm working from H.M. Schey's Div, grad, curl, and all that, and am trying to figure out surface integration.

One of the example problems boils down to the following surface integral over a projection, with z = 1 - x - y
$$\sqrt{3} \int \!\!\! \int_R 1 - y \,dx \,dy$$
I made x and y go from 0 to 1.
$$\sqrt{3} \int_0^1 \!\!\! \int_0^1 1 - y \,dx \,dy$$
According to the book, the answer is $$\frac{1}{\sqrt{3}}$$, but I keep getting $$\frac{\sqrt{3}}{2}$$.

Another example problem has z = 1 - y - x/2:
$$\int \!\!\! \int_R \frac{3x}{4} - \frac{3y}{2} + \frac{1}{2} \,dx \,dy$$
I made x go from 0 to 2, and y go from 0 to 1.
$$\int_0^1 \!\!\! \int_0^2 \frac{3x}{4} - \frac{3y}{2} + \frac{1}{2} \,dx \,dy$$
The answer is supposed to be 1/2, but I keep getting 1.

I have not completely lost faith, because I am able to evaluate the integrals which require conversion to polar coordinates. However, the ones in cartesian coordinates are confusing me.

2. Oct 1, 2006

### HallsofIvy

Okay, why did you "make x and y go from 0 to 1" and what happened to x in the integrand? It's hard to know if that right or wrong or if your integrand is correct since you don't tell us what the problem says!

Please give the problems themselves so we can see if what you are doing is right or wrong.

3. Oct 1, 2006

### Saketh

I knew I was forgetting something when I wrote the above post...sorry!

Here are the two problems:
"Let's first compute the surface integral
$$\int \!\!\! \int_S (x + z) \,dS$$
where $S$ is the portion of the plane $x + y + z = 1$ in the first octant.
"

The second one:
"Suppose we wish to calculate $$\int \!\!\! \int_S \vec{F} \cdot \hat{n} \,dS$$ where $$\vec{F}(x, y, z) = \hat{i}z - \hat{j}y + \hat{k}x$$, and S is the portion of the plane
$$x + 2y + 2z = 2$$
bounded by the coordinate planes.
"

4. Oct 2, 2006

### Saketh

I made x and y go from 0 to 1 because the projection R of the surface S on the xy-plane extends from x = 0 to x = 1, and y = 0 to y = 1.

5. Oct 2, 2006

### HallsofIvy

Not for all x and y! The projection of the plane onto the first quadrant of the xy-plane (since the question asks about the first octant) is a triangle with borders x= 0, y= 0, and x+ y= 1. The limits of integration for a double integral are all constants only if the region is a rectangle. x can go between 0 and 1 but for each x y varies between 0 and 1- x. The integral should be
$$\int_{x=0}^1\int_{y= 0}^{1-x} (x+z)dV$$.

A vector perpendicular to x+ y+ z= 1 is grad(x+ y+ z)= i+ j+ k which has length $\sqrt{3}$ so $dV= \sqrt{3}dxdy$. Of course, on x+ y+ z= 1, z= 1- x- y so x+ z= x+ 1- x- y= 1- y. Your integral is
$$\sqrt{3}\int_{x=0}^1\int_{y= 0}^{1- x}(1- y)dy dx$$.

For your second problem, assuming that the question was to integrate on the portion of z = 1 - y - x/2 in the first octant, the plane cuts the xy-plane in the line 0= 1- y- x/2 or y= 1- x/2 and the projection of the plane onto the first quadrant of the xy-plane is a triangle with vertices at (0,0), (0, 1), and (2, 0). If you let x vary from 0 to 2 then for each x, y varies from 0 to 1- x/2. The integrals will be
$$\int_{x= 0}^2\int_{y= 0}^{1-\frac{x}{2}}( ) dydx[/itex] If you don't like fractions, you can reverse the order of integration, letting y vary from 0 to 1 and, for each y, x varies from 0 to 2y- 2: [tex]\int_{y=0}^1\int_{x= 0}^{2y- 2}( )dxdy$$

6. Oct 2, 2006

### Saketh

I did not use a grad (what is a grad?) to find the length of the normal vector - I found what the normal vector is in terms of a function, namely this:
$$\hat{n} = \frac{ -\hat{i}\frac{\partial f}{\partial x} - \hat{j}\frac{\partial f}{\partial y} + \hat{k} } { \sqrt{ 1 + \left ( \frac{\partial f}{\partial x} \right )^2 + \left ( \frac{\partial f}{\partial y} \right )^2 } }$$.

I somewhat understand your explanation, HallsOfIvy, but I am still confused. In the first problem, how can you know that x goes from 0 to 1, but have to write that y goes from 0 to 1-x?

Or is it that you are writing y as a function of x so that the integral finds the area correctly?