Surface Area of Plane Inside Cylinder: Solved

[adm_strat]

[SOLVED] Surface Area

Homework Statement

Find the area of the surface of the part of the plane x + 2y + z = 4 that lies inside of the cylinder x^{2} + y^{2}=4

Homework Equations

A(S)= \int\int_{D} \sqrt{1+( \frac{\partial z}{\partial x})^{2} + +( \frac{\partial z}{\partial y})^{2}} dA

The Attempt at a Solution

I can tell intuitively that the intersection is a ellipse. When I set the two equations equal to each other I get the equation:

z= x^{2}-x+y^{2}-2y

I am having a brain fart and can't seem to do the double integral. Sorry, its finals week. I know that I need to do a change of variables, but I don't know what since it is an ellipse in R^{3} I can complete the square, but it didn't seem to lead me anywhere useful.

Any help would be appreciated. Thanks in advance.

 

[coomast]

I assume the equation of the plane is:

x+2y+z=4

So, what is the meaning of the symbols in the integral you gave (it is the correct one for calculating the area)? What is dA and what is the exact meaning of z? In looking back to their meaning, you will notice that you do not need to set the two equations equal to each other. The equation of the plane is related to the z value and the cylinder is related to dA. Once you see this, you can set up the integral and calculate it. Consider transformation to cylinder coordinates, it will make things a lot easier.

 

[HallsofIvy]

Because you are integrationg over the base of a cylinder, you might want to put the integral in polar coordinates.

 

[coomast]

I assume the equation of the plane is:

x+2y+z=4

So, what is the meaning of the symbols in the integral you gave (it is the correct one for calculating the area)? What is dA and what is the exact meaning of z? In looking back to their meaning, you will notice that you do not need to set the two equations equal to each other. The equation of the plane is related to the z value and the cylinder is related to dA. Once you see this, you can set up the integral and calculate it. Consider transformation to cylinder coordinates, it will make things a lot easier.

Because you are integrationg over the base of a cylinder, you might want to put the integral in polar coordinates.

Indeed HallsofIvy, I meant polar...