# Surface Area and Surface Integrals

1. Nov 8, 2007

### mit_hacker

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

(Q) Find the area of the surface cut from the paraboloid x^2+y+z^2 = 2 by the plane y=0.

2. Relevant equations

3. The attempt at a solution

The unit normal vector in this case will be j. Moreover, the gradient vector will be
sqrt(4x^2+4z^2+1). And the denominator which is the dot product of the gradient vector and j is 1 so we need not bother about that.

So the double integral will be that of sqrt (4x^2+4z^2+1) but since y=0, it means that x^2+z^2 = 2 so sqrt (4x^2+4z^2+1) becomes 3.

The problem is that the solution manual does not do this. It retain the integral as sqrt (4x^2+4z^2+1) and uses polar co-ordinates which I can do but I don't understand why we cannot substitute. Please explain.

Thank-you very much for the time and effort!!!

2. Nov 9, 2007

### HallsofIvy

Staff Emeritus
Your first statement is wrong! In fact, after reading that "the unit normal vector in this case will be j", my first reaction was "unit normal vector to what?", certainly not the paraboloid! Then I thought, "No, it's to the y= 0 plane", and assumed you meant the solid bounded by the paraboloid and the area cut off that by the plane. But that's trivially a disk with area 2$\pi$ so that's not what's meant.
Now, rereading the problem, we're back to my original reaction: the "surface" involved is the part of the paraboloid x2+ y+ z2= 2 above the plane y= 0. The "normal" needed is the normal to that surface. A standard way to do that is this: think of z2+ y+ z2= 2 as y= 2- x2- z2, in terms of the two parameters x and z. In vector form that is $\vec{r}= x\vec{i}+ (2- x^2- z^2)\vec{j}+ z\vec{k}$. Then $\vec{r}_x= \vec{i}- 2x\vec{j}$ and $\vec{r}_z= -2z\vec{j}+ \vec{k}$ and the "fundamental vector product" is the cross product of those vectors: $-2x\vec{i}-\vec{j}-2z\vec{k}$. That vector is normal to the paraboloid at each point and its length, $\sqrt{4x^2+ 4z^2+ 1}$,times dxdz, is the differential of surface area, as your book says.

Last edited: Nov 9, 2007
3. Nov 9, 2007

### mit_hacker

Thanks a ton!!!

Thanks a lot for your help. I understand everything clearly now. The problem was that my book shows an extremely vague and misleading solution to approach such problems which was throwing me all around.

Thanks a ton once again!