Surface Area of a part of a plane inside an ellipsoid.

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SUMMARY

The surface area of the part of the plane defined by the equation 9x + 10y + z = 6, which lies inside the elliptic cylinder described by the equation x²/25 + y²/100 = 1, is calculated using a double integral approach. The correct surface area is determined to be 50π√182. The calculation involves parameterizing the plane and using the formula for the differential area element dS, resulting in the integral ∫∫_A (1)√182 dx dy, where the area A corresponds to the area of the ellipse, calculated as π(5)(10).

PREREQUISITES
  • Understanding of surface integrals and parameterization of surfaces.
  • Familiarity with elliptic cylinders and their equations.
  • Knowledge of vector calculus, specifically Stokes' Theorem.
  • Ability to compute double integrals and area calculations for ellipses.
NEXT STEPS
  • Study the application of Stokes' Theorem in surface integrals.
  • Learn about parameterization techniques for different surfaces in vector calculus.
  • Explore the properties and calculations of elliptic cylinders in three-dimensional space.
  • Practice solving double integrals involving non-rectangular regions.
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Students and professionals in mathematics, particularly those studying calculus and vector analysis, as well as anyone involved in engineering fields requiring surface area calculations in three-dimensional geometry.

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Homework Statement



Find the surface area of that part of the plane 9x+10y+z=6 that lies inside the elliptic cylinder \frac{x^2}{25} +\frac{y^2}{100} =1



2. The attempt at a solution

Once again I was just told that the surface area would be equal to the double integral of the area of the ellipse times the normal vector of the plane. Which gives me the correct answer being 50pi\sqrt{182} but I have no clue how this was obtained. I'm looking at my book for answers, for equivalencies in Stokes' Theorem that would indicate this but I can't find anything.

Is this because to calculate a surface integral, we must approximate the patch area of S and in this case we can actually find the area rather than using the cross product of the partials of the vector?
 
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If you parameterize the plane as

R(x,y) = <x, y, 6-9x-10y>

and use the formula dS = |Rx X Ry|dx dy

you get dS = \sqrt{182}\, dxdy, and the area becomes

\int\int_A (1)\sqrt{182}\, dxdy

which is \sqrt{182}Area(A) and, of course, the area of the ellipse is \pi(5)(10)
 

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