Double integral, find volume of solid

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SUMMARY

The discussion focuses on calculating the volume of a solid defined by the parabolic cylinders \(y = 1 - x^2\) and \(y = x^2 - 1\), along with the planes \(x + y + z = 2\) and \(4x + 5y - z + 20 = 0\). The correct approach involves determining the limits of integration for \(x\) and \(y\) by graphing the parabolas, which confirms that \(x\) should range from -1 to 1. The final volume calculation using the double integral \(\int_{-1}^{1} \int_{x^2-1}^{1-x^2} (5x + 6y + 18) \, dy \, dx\) yields the correct volume of \(\frac{53}{2}\).

PREREQUISITES
  • Understanding of double integrals and volume calculations
  • Familiarity with parabolic equations and their graphs
  • Knowledge of plane equations and solving for \(z\)
  • Proficiency in using integration techniques in calculus
NEXT STEPS
  • Graph the equations \(y = 1 - x^2\) and \(y = x^2 - 1\) to visualize the region of integration
  • Practice solving double integrals with varying limits of integration
  • Explore applications of double integrals in calculating volumes of solids
  • Review the method of subtracting volumes defined by different surfaces
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable integration, as well as educators teaching volume calculations using double integrals.

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Homework Statement


Find the volume of the solid by subtracting two volumes, the solid enclosed by the parabolic cylinders:
y = 1 − x2,
y = x2 − 1
and the planes:
x + y + z = 2
4x + 5y − z + 20 = 0

Homework Equations


∫∫f(x,y) dA

The Attempt at a Solution



So I solved for z in the plane equations:
z=2-x-y
z=4x+5y+20

I subtracted these two equations:
(4x+5y+20)-(2-x-y) = 5x+6y+18 = z

01x2-11-x2 5x+6y+18 dy dx

=53/2

It's the wrong answer, and I think my x boundaries might be between -1 and 1 after graphing it but I'm not sure.
 
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If you graph ##y=1-x^2## and ##y=x^2-1## in the xy plane that should settle the ##x## limits for you. If you use ##x=-1## for the lower limit, does that fix it for you?
 
LCKurtz said:
If you graph ##y=1-x^2## and ##y=x^2-1## in the xy plane that should settle the ##x## limits for you. If you use ##x=-1## for the lower limit, does that fix it for you?
Yes, it's correct.
 

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