Triple Integrals with Cylindrical Coordinates

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SUMMARY

The discussion focuses on evaluating the triple integral of the function 2(x^3 + xy^2) over the solid E in the first octant beneath the paraboloid z = 9 - x² - y². The correct bounds for the integral should be θ from 0 to π/2, r from 0 to 3, and z from 0 to (9 - r²). The user initially set θ from 0 to 2π, leading to an incorrect result of 0, which was identified as a mistake in the bounds for the first octant.

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Familiarity with cylindrical coordinates (x = rcosθ, y = rsinθ)
  • Knowledge of evaluating integrals over specified regions
  • Ability to interpret geometric shapes like paraboloids
NEXT STEPS
  • Study the evaluation of triple integrals in cylindrical coordinates
  • Learn about the geometric interpretation of paraboloids and their equations
  • Practice setting correct bounds for integrals in different octants
  • Explore common mistakes in integral calculus and how to avoid them
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and integral evaluation techniques.

dancingmonkey
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Homework Statement


Evaluate the integral, where E is the solid in the first octant that lies beneath the paraboloid z = 9 - x2 - y2.

∫∫∫(2(x^3+xy^2))dV

Homework Equations



x=rcosθ
y=rsinθ
x^2+y^2=r^2

The Attempt at a Solution



θ=0 to 2π, r=0 to 3, z=0 to (9-r^2)

2(x^3+xy^2)=2x(x^2+y^2)=2rcos(θ)(r^2)

∫0 to 2π ∫0 to 3 ∫0 to (9-r^2) (2rcos(θ)r^2)rdzdrdθ

I was wondering if my bounds were correct. And when I solved the integral I keep getting an answer of 0, which is incorrect. Can someone please help me with this problem?
 
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If you're integrating over the first octant, then theta doesn't go from 0 to 2pi. Other than that, the rest is ok.
 
Thank you so much! That was the problem. I missed the part where it said the first octant.
 

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