Surface Area and surface Integrals

[mit_hacker]

Homework Statement

(Q) Find the area of the portion of the surface x^2 - 2z = 0 that lies above the triangle bounded by the lines x = sqrt(3), y = 0, and y = x in the xy-plane.

Homework Equations

The Attempt at a Solution

The know how to find the gradient vector. The part where I am facing a problem is in finding the unit normal vector. the solution manual simply states that it is equal to k.

However, I thought that since the surface is defined by an equation that contains only x and z, the normal vector must be j.

I've been facing lots of problems in finding the normal vector lately. Can someone please explain to me the procedure to do so?

Thank-you very much!

 

[HallsofIvy]

Homework Statement

(Q) Find the area of the portion of the surface x^2 - 2z = 0 that lies above the triangle bounded by the lines x = sqrt(3), y = 0, and y = x in the xy-plane.

Homework Equations

The Attempt at a Solution

The know how to find the gradient vector. The part where I am facing a problem is in finding the unit normal vector. the solution manual simply states that it is equal to k.

However, I thought that since the surface is defined by an equation that contains only x and z, the normal vector must be j.

I've been facing lots of problems in finding the normal vector lately. Can someone please explain to me the procedure to do so?

Thank-you very much!

The surface is given by x^2- 2z= 0 or z= (1/2)x^2, a "parabolic cylinder" with axis along the y-axis. No, its "normal vector" is NOT j. it should be obvious that, because the surface is not a plane, its normal vector changes direction from point to point.
Write the surface as the vector equation \vec{r}= x\vec{i}+ y\vec{j}+ (1/2)x^2\vec{k}. Then \vec{r}_x= \vec{i}+ x\vec{k} and \vec{r}_y= \vec{j}. Those are two vectors in the tangent plane to the surface at each point and so their cross product (the "fundamental vector product") is normal to the surface: -x\vec{i}+ \vec{k}. The differential of area is the length of that times dxdy:
\sqrt{x^2+1}dxdy.
(You keep this up and you are going to give MIT a bad name!)

 

[mit_hacker]

Thanks again!

I get your solution!

And between you and me, I'm not actually an MIT student but hope to be so I'm trying hard to be good. Thanks to people like you, I'll get the support I need!

 

[mit_hacker]

But...

I understand your approach to finding the Area but how will we use that method when there are two values of x specified. To be more clear, how will you use that method in the following question:

Find the area of the region cut from the plane x+2y+2z=5 by the cylinder whose walls are x=y^2 and x=2-y^2.

 

[HallsofIvy]

That "cylinder" cuts off a region in the xy-plane. I would first get the "differential of area" by writing the plane as z= (5/2)-(1/2)x- y so the "position vector" is \vec{r}= x\vec{i}+ y\vec{j}+ (5/2)-(1/2)x- y)\vec{k}. Then \vec{r}_y= \vec{j}-\vec{k} and \vec{r}_x= \vec{i}-(1/2)\vec{k}. The "fundamental vector product" is their cross product and so a normal vector is -(1/2)\vec{i}+\vec{j}-\vec{k}. (That's a constant precisely because this is a plane.) Its length is 3/2 so the "differential of area" is (3/2)dxdy. (I chose to solve for z and so write the plane in terms of "parameters" x and y because the "cylinder" was parallel to the z-axis.)

Now, you only need to determine the region of integration: it is the region between x=y² and y= x² and those intersect at (0,0) and (1,1). If you want to make the outer integral with respect to x, x must range from its lowest value in that region, 0, to its highest, 1. For each x, then, y must range from the lower parabola, y= x², to the upper, x=y² or y= \sqrt{x}. Your surface area will be simply
\int_{x=0}^1\int_{y= x^2}^{\sqrt{x}} \frac{3}{2} dydx
Of course that is simply 3/2 times the area of the region in the xy-plane- tilting a plane simply changes the area cut off.