Question

Solve the surface integral: \[\int\limits_{}^{}\int\limits_{S}^{}xz \ dS\] S is the part of the plane 2x+2y+z=4 that lies in the first octant

Answer 1

I don't even know where to begin with this. I think you have to find r_u and r_v to start but I don't know how to find those. @TuringTest @UnkleRhaukus @eseidl @experimentX @phi

Answer 2

Ok so I found the first step, but I'm not sure how they got those points or those vectors

Answer 3

First you have to parametrize the surface in terms of x and y in this case. you can solve the equation of the plane for z and that puts the plane in terms of x and y.

so we get the parametrized surface in vector format to be:
\[\Phi(x,y)=\left( x,y, g(x,y) \right) \]
which becomes:
\[\Phi(x,y)=\left( x, y, 4-2x-2y \right)\]

Answer 4

Then you have to find the tangent vector to the parametrized surface and take the cross product of those to get the normal vector.

Answer 5

I got:
\[T _{x}=(1,0,-2) \]
\[T _{y}=(0, 1, -2)\]

Answer 6

This seems a lot more familiar, but it doesn't match up with what the answer key showed?

Answer 7

The goal is to get

\[\int\limits_{?}^{?}\int\limits_{?}^{?}_{D} f(x, y, g(x,y)\left| n(x,y) \right|\]

Answer 8

Would be it:
\[\int\limits\limits_{?}^{?}\int\limits\limits_{?}^{?}x*(4-2x-2y)*3 \ dS\]

Answer 9

I get the normal vector to be
\[n=(2, 2, 1)\]

then taking its magnitude I get:

\[\sqrt{2^{2}+2^{2}+1^{2}}=\sqrt{5}\]

Answer 10

Yeah, but you found r_x = (1,0,-2) and r_y=(0,1,-2) so the magnitude of the cross product would be |2i+2j+1k|=sqrt(9)=3
Where did n come from?

Answer 11

Nevermind your n is the cross product, the only mistake is that the magnitude is 3

Answer 12

yes true.....sorry bad arithmetic

Answer 13

I'm confused as to what comes next

Answer 14

ok i think it should be
\[\int\limits_{0}^{2}\int\limits_{0}^{2} 3x(4-2x-2y) dxdy \]

bear with me, I just learned these a couple weeks ago and maybe a bit shaky.....

Answer 15

I completely understand. I can't stand these!

Ok, I got the same thing you did for the function, but how did you find the bounds?

Answer 16

yeah they suck!!! I have a test coming up in this soon too....

Since we are finding they told us the surface is in the first octant I set y and z = 0 and solved for x and then did the same for y. If you draw out the plane you will see where the plane intersects the x and y axis is 2

Answer 17

Got it! Thank you! You have no idea how long I've been stuck on this problem!!

Answer 18

did we get the correct answer using that method?

Answer 19

I'll check with WA

Answer 20

cool let me know, I hope I didn't lead you astray, and I have the complete idea how long you've been stuck!!! trust me....I'm in the same boat! Calc 3 is a bear!

Answer 21

Ok, so wolfram alpha says it is -8 and the answer from the site I found says the answer is 20, but someone commented saying that they found the answer as 4, so I'm not sure what to think. Everything looks correct though so I think I'm going to stick with -8!

Thanks again and best of luck on your test!

Answer 22

I also got -8 for the answer. Good luck with your class!!!!

Answer 23

Thanks!

Answer 24

This may not be of interest, but here is a write-up on doing this problem