Question

Question about surface area.

Answer 1

@ganeshie8 @dan815 @Jhannybean

Answer 2

Ok and please don't spam questions here x_x I was so confused last time haha

Answer 3

Pretty sure it's very simple, but i'm brain dead now after all teh discussion... hahaha gotchya... lolol

Answer 4

Well first off, if it's a rectangle you can use rectangular coordinates!

Answer 5

please, your rectangle lies on the plane whose equation is:
-y+z=0

Answer 6

Haha really :(

I based it off the coordinates lol.

Answer 7

I think that your answer is:
\[\int\limits_{0}^{1}xdx \int\limits_{0}^{1}zdz \int\limits_{0}^{z}ydy\]

Answer 8

Oh, I was right!! :(

Answer 9

please @Jhannybean note that coordinates x, y, z are not independent, they are constrained by the plane

Answer 10

Oh you are right, I did not see your integral properly! i noticed the upper limit on the y.

Answer 11

x = u
y = v
z = v

||Tu x Tv|| = ||(1,0,0) x (0, 1, 1)|| = sqrt(2)

rest should be easy

Answer 12

nevertheless I think my answer is wrong, because the right answer is a surface integral, dS is the surface differential element.
my answer is the measure of function f(x,y,z) on a plane -y+z=0, instead!

Answer 13

\[\int_R xyz dS = \sqrt{2}\int\limits_{..}^{..} \int\limits_{..}^{..} uv^2 dudv\]

Answer 14

yeah finding the bounds can be tricky ^

Answer 15

u : 0->1
v : 0->1
should work, double check..