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Surface Area 1

  • Thread starter rocomath
  • Start date
1,750
1
The part of the plane 3x+2y+z=6 that lies in the first octant.

[tex]A(s)=\int_0^2\int_0^3\sqrt{14}dydy[/tex]

Are my limits not correct? B/c my answer is just off by a little.

me: 6\sqr(t14), answer: 3\sqrt(14)
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
No, not correct. The domain of integration is a triangle in the x-y plane, not a rectangle. So the y limits should depend on x (or vice versa).
 
1,750
1
No, not correct. The domain of integration is a triangle in the x-y plane, not a rectangle. So the y limits should depend on x (or vice versa).
dope! Gotcha, thanks.
 

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