Confused on double integral in polar cords

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SUMMARY

The discussion focuses on calculating the volume of a solid enclosed by the hyperboloid defined by the equation -x² - y² + z² = 1 and the plane z = 2 using polar coordinates. The user correctly identifies the intersection curve, leading to the integral setup of ∫₀²π ∫₀√3 (1 + r²)^(1/2) r dr dθ. However, the user mistakenly calculates the volume by not properly applying the upper and lower bounds for z, resulting in an incorrect answer of 14/3 π instead of the correct volume of 4/3 π. The key takeaway is the importance of correctly identifying zupper and zlower in volume integrals.

PREREQUISITES
  • Understanding of polar coordinates in multivariable calculus
  • Familiarity with hyperboloid equations
  • Knowledge of double integrals and volume calculations
  • Ability to perform substitutions in integrals
NEXT STEPS
  • Review the concept of volume calculation using double integrals in polar coordinates
  • Study the properties and equations of hyperboloids
  • Learn about the correct application of upper and lower bounds in triple integrals
  • Practice solving similar problems involving polar coordinates and volume
USEFUL FOR

Students studying multivariable calculus, educators teaching integration techniques, and anyone looking to improve their understanding of volume calculations in polar coordinates.

Samuelb88
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Homework Statement


Use polar coordinates to find the volume of the solid enclosed by the hyperboloid -x^2-y^2+z^2=1 and the plane z=2.

The Attempt at a Solution


Solving for z of the equation of the hyperboloid I find z = Sqrt(1 + x^2 + y^2). Letting z = 2 to determine the curve of intersection I find that 3 = x^2 + y^2, or r = Sqrt(3). Thus:

\int _{0}^{2Pi} \int _{0}^{3^(^1^/^2^)} (1+r^2)^(^1^/^2^)rdrd\theta

Making the substitution u = 1 + r^2 gives:

\frac{1}{3}\right) \int _{0}^{2Pi} 7d\theta = \frac{14}{3} Pi

The back of my book has 4/3*Pi. I don't understand how I am doing this problem wrong.
 
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Your integrand should be zupper - zlower.
 

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