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So, going through triple integral volume problems, when establishing the bounds of integration for a solid bound by a paraboloid, one common trick is to convert to cylindrical coordinates allowing for the substitution of x^2+y^2 to r^2. However, what if the coefficient values in front of x and y are different? If for example the equation for the paraboloid that binds the solid is z=x^2/36 + y^2/100? Would I have to set up the triple integral in Cartesian coordinates? I set up my triple integral but it looks exceedingly difficult. How else could I approach this problem.

Math Loser
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  • For some value $a$, write a^b to generate $a^b$. In your case, $a = r,x,y,z$ and $b=2$. However, add dollar signs before and after your expresdion or equation, i.e. $$$$\ldots$$$$. Thus, $$$a^b$$$ makes $a^b$. – Mr Pie Apr 04 '18 at 06:28
  • @MathLoser Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 07 '18 at 19:53

1 Answers1

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HINT

Let use

  • $x=6r\cos \theta$
  • $y=10r\sin \theta$
  • $z=z$

calculating the jacobian accordingly.

user
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