Question

Surface area of the plane: 2x+y+2z=16;
Bounded by x^2+y^2=64; and x,y>=0 (first quadrant)

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

The surface area of a function is defined as:
\[\int\limits \int\limits \sqrt{(f_x)^2+(f_y)^2+1}dA\]
From here, you can use the circle you are given to find your bounds for x and y.
For convenience I would convert it to polar coordinates since you're dealing with part of a circle :P Let me know if you get stuck!

Answer 3

Also, if you switch to polar. Don't forget:
\[dA=rdrd \theta\]

Answer 4

Ok i used a different definition but still get to this same point:
\[3/2\int\limits_{0}^{\pi/2}\int\limits_{0}^{8}rdrd \Theta\]
Solving from there i get 24*pi, which is not the correct answer (answer: 192/5+48sin-1(3/5)

Answer 5

I get this: \[\sqrt{6}\int\limits_{0}^{\pi/2} \int\limits_{0}^{8}r dr d \theta\]

Answer 6

I'm not sure if that gives you whatever they got, but it may be a start? x.x

Answer 7

If you can't figure it out I'll see what I can do tomorrow, I need to get to sleep I have an exam tomorrow D: