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I have to evaluate the $\iint_S z ds$ where $S$ is the part of the of the plane $5x+3y+z=15$ that lies in the first octant.

I have been working on this problem for lon capa but I keep getting the wrong answer. I just want to make sure that my integral is correct so I can figure out if my error is in my computation or if it is in the beginning of the problem.

I'm taking the integral $$ \int_0^3 \int_0^{5-(5/3)x} (15-5x-3y)\sqrt{35} \ \ dy \ dx $$

Okay I first integrated with respect to y and got

$$ \int_0^{5-(5/3)x} \ 15y-5xy - \frac{3}{2}y^2 \ \ dy $$ Solving this I got $ \ \frac{75}{2}-25x-\frac{25}{6}x^2 \ . $ Then I integrated this from 0 to 3 and got $ \ \frac{-75}{2} \ (\sqrt{35}) \ . $

colormegone
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Ayoshna
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  • The integral looks alright. What did you do from there? Why do you believe your result is wrong? – colormegone Apr 10 '14 at 00:27
  • Parameterise using $\alpha(x,y) = (x,y, 15-5x-3y)$ and $V(D\alpha) = \sqrt{(1+5^2)(1+3^2) - 15^2} = \sqrt{35}$. So your integrand is correct. You then want to integrate $y$ from $0$ to $(15-5x)/3$. So, yes the answer is right. – Christopher K Apr 10 '14 at 00:31
  • I solved the problem and plugged it into lon capa and it told me that the answer was incorrect so I must be solving it wrong. Let me edit my question and show my work. – Ayoshna Apr 10 '14 at 09:45
  • I wrote out the intermediate steps and it looks like you have a sign error in one term. (Reading your post was a little misleading, since it appears that you made the calculation mostly correctly, but you didn't quite write what you actually did.) – colormegone Apr 10 '14 at 19:04

1 Answers1

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Your double integral is correct, but there appears to be a small error in the execution. Your next steps should be

$$ \int_0^3 \int_0^{5-(5/3)x} (15-5x-3y) \ \sqrt{35} \ \ dy \ dx \ \ = \ \ \sqrt{35} \ \int_0^3 \left( \ 15y-5xy - \frac{3}{2}y^2 \ \right) \ \vert_0^{5-(5/3)x} \ \ dx $$

$$ = \ \ \sqrt{35} \ \int_0^3 \ \left[ \ (15 -5x) \ (5-\frac{5}{3}x) \ - \ \frac{3}{2} (5-\frac{5}{3}x)^2 \ \right] \ \ dx $$

$$ = \ \ \sqrt{35} \ \int_0^3 \ \left[ \ (75 -50x + \frac{25}{3}x^2) \ - \ (\frac{75}{2} - 25 x + \frac{25}{6}x^2) \ \right] \ \ dx $$

$$ = \ \ \sqrt{35} \ \int_0^3 \ \frac{75}{2} \ - \ 25 x \ + \ \frac{25}{6}x^2 \ \ dx \ \ . $$

Try it from this point. The result should be a positive number, since you are integrating $ \ z \ $ over a surface in the first octant.

colormegone
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