Finding Surface Area in the First Octant and Left of a Given Plane

[Infernorage]

How would I find the area of the surface z=2-2x^(3/2) in the first octant and to the left of the plane x+y=1? Could someone solve this and explain to me how to do it, because I really am unsure of what to do. Thanks in advance.

 

[Office_Shredder]

You probably need to do some sort of integration. What do you think?

 

[Infernorage]

You probably need to do some sort of integration. What do you think?

Lol yea I got that. What I don't understand is how to get the equation that needs to be integrated and the limits.

 

[Matthollyw00d]

Try drawing it. Even when in R^3, sketches still help tremendously in these cases

 

[HallsofIvy]

In other words, they are suggesting that you try yourself and show what you have tried. There are many suggestions we could make, but what you will understand depends on what you already know about this kind of problem and we do not yet know that.

 

[Infernorage]

Okay, well I found the y and x partial derivatives and plugged them into the surface area equation and got \int \int \sqrt{9x+1}. The main thing I am confused about is how to find the limits. Do I have to find the intersection of the surface z=2-2x^(3/2) and x+y=1? Can someone give me a quick reminder of how to find that intersection? Thanks.

 

[HallsofIvy]

Here's a way that helps me think of these: draw an xz- coordinate system and sketch z= 2- 2x^{3/2} (in the first quadrant, it looks a bit like a parabola). Draw a separate xy-coordinate system and sketch x+ y= 1. Now imagine moving the xz-coordinate graph onto of the xy-coordinat graph so that the z-axis is straight up. z= 2-2x^2 is a "cylinder" and x+y= 1 is a plane going straight up. To cover the area you want over the xy-plane, x must go from 0 to 1 and, for each x, y must go from 0 to y= 1- x (from solving x+y= 1 for y). Those are your limits of integration.