Question

What's the area of the surface described by the function\[z=f(x,y)=x^2-y^2-xy\]in the region\[R=[-1,1]\times[-1,1]?\]Go, go, go! :)

Answer 1

What are you doing? Typing a message to yourself?

Answer 2

I'm surprised you don't already know the answer to this

Answer 3

Yeah, I don't know it. :)

Answer 4

Satellite is here so ask him

Answer 5

Solving this involves double integrals

Answer 6

yeah... i just wanted to see someone use ∯.

Answer 7

i think i should ask purely algebraic questions n_n

Answer 8

\[{\int\limits \int\limits}_R \sqrt{\frac{\delta g}{\delta x} + \frac{\delta g}{\delta y}^2} \]

Answer 9

I don't know how how to make that symbol

Answer 10

forgot dxdy

Answer 11

Wasn't it\[\iint_{D}g\left(x,y,f\left(x,y\right)\right)\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2}dA\]

Answer 12

what is g?

Answer 13

Yeah, so it is clear that you already know how to set it up

Answer 14

g = x^2 - y^2 - xy

Answer 15

It's a function of the independent and dependent variables, iirc.

Answer 16

if you are just looking for the surface area then you want

\[\iint\limits_{R}\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}\,\,\,dA\]

Answer 17

I think there's a mistake in the above expression; it should be\[\iint_{D}g\left(x,y,f\left(x,y\right)\right)\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}dA\]

Answer 18

(I meant in mine xd)

Answer 19

then you are not looking for the surface area...you are computing the scalar function over a surface

Answer 20

that formula you have is a little mixed up

Answer 21

I see what you're saying now.

Answer 22

\[\iint_{D}f\left(x,y,g\left(x,y\right)\right)\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}dxdy\]

Answer 23

Then by letting\[f\left(x,y,g\left(x,y\right)\right)=1\]you're computing just the surface area?

Answer 24

I'm not here anymore

Answer 25

yes...the surface area over g

Answer 26

Hero, I tend to disguise questions to refresh myself of old, dreaded subjects which I need to be aware of for the classes I'm currently taking. xd

I attached an excerpt.