Question

Evaluate the surface integral over S of the function yz, where S is the part of the plane x + y + z = 1 that lies in the first octant. I can't seem to reach the correct answer of: the square root of 3 devided by 24.

Answer 1

In what order are you integrating the variables and what are your limits of integration?

Answer 2

I first tried the order of dxdy and then I tried dydx.... The limits with the type 1 area were \[0 \le x \le 1\] and \[0 \le y \le -x + 1\]. I don't know if my mistake lies with the limits, but I can't seem to reach the correct answer.

Answer 3

I started by paramatrizing the surface as r( x, y ) = < x, y, 1-x - y >. Then \[dS = \sqrt{1 + (Z _{x})^{2} + (Z _{y})^{2}} dA = \sqrt{3} dA\]. Then \[\int\limits_{?}^{?}\int\limits_{?}^{?}_{S} yz dS = \int\limits_{0}^{1}\int\limits_{0}^{1 - y} y( 1 - x - y ) \sqrt{3}dxdy\] and then I reach the answer \[-7\sqrt{3} / 12\]

Answer 4

Your first couple of steps look dead on. What do you get after the first integration?

Answer 5

I have as an integrand
\[ y^3/2 - y^2 + y/2 \]

Answer 6

sqrt(3) times that

Answer 7

Then integrating that from 0 to 1, I end up with
\[ \frac{1}{\sqrt{3}} ( 1/8 - 1/3 + 1/4 ) \]
which gives the right answer.

Answer 8

\[\sqrt{3}\int\limits_{0}^{1}(y ^{3} - 5/2y ^{2} +2y - 1/2) dy\]

Answer 9

Ok, so radically different. Let me write out that inner integral as I see it ...

Answer 10

\[ \int_0^{1-y} y(1−x−y) dx = \left[ yx - x^2y/2 - y^2x \right]_0^{1-y} \]
\[ = y(1-y) - (1-y)^2y/2 - y^2(1-y) \]
\[ = y^3/2 - y^2 + y/2 \]

Answer 11

make sense?

Answer 12

Yes that makes perfectly sense... My mistake whas with the second term, I didn't multiply it with the y.... Thanks a lot, now I got it right...

Answer 13

Great.

Answer 14

And thanks for the problem. I haven't done a surface integral for ages and it's good to calculate it again!

Answer 15

My pleasure... Stay tuned there may be more coming your way :P