What is the volume of the solid bounded by two paraboloids?

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Discussion Overview

The discussion revolves around finding the volume of a solid bounded by two paraboloids, specifically z=x^2+y^2 and z=8-x^2-y^2. Participants explore the region of integration and the limits necessary for volume calculation, considering different methods of integration.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant describes the properties of the two paraboloids, noting their vertices, orientations, and the nature of their cross sections.
  • Another participant identifies the intersection of the two paraboloids, determining that it occurs at the circle defined by x^2 + y^2 = 4 in the z=4 plane.
  • Multiple participants inquire about the appropriate function to integrate for volume calculation, with some suggesting the use of a triple integral and others advocating for a double integral approach.
  • There is a suggestion to use polar coordinates for integration, with specific limits provided for both Cartesian and polar forms.
  • Some participants express confusion about whether to start with a triple integral or a double integral, indicating differing interpretations of the problem's requirements.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method of integration to use, with some favoring a triple integral approach while others insist on a double integral. The discussion remains unresolved regarding the best approach to calculate the volume.

Contextual Notes

There are indications of missing assumptions regarding the integration method, and the discussion reflects varying interpretations of the instructions provided for the problem.

cbarker1
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Hello Everyone,

I need to find the volume of this solid bounded by the paraboloids z=x^2+y^2 and z=8-x^2-y^2. I need to find the region of integration. I need to find the limits of integration as well. I tried to graph the two surfaces. Thanks

Cbarker1
 
Last edited:
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I presume you saw that the first is a paraboloid with vertex at (0, 0, 0), opening upward, with z-axis as its axis, and every cross section at given z a circle with center at (0, 0, z) and radius \sqrt{z}. The second is a paraboloid with vertex at (0, 0, 8), opening downward, with z-axis as it axis, and every cross section is a circle with center at (0, 0, z) and radius \sqrt{8- z}. The two paraboloids will intersect where z= x^2+ y^2= 8- x^2+ y^2 which reduces to x^2+ y^2= 4 so the intersection is the circle with center (0, 0, 4), radius 2, in the z= 4 plane.

Since both paraboloids open outward to that circle, you can cover the whole figure by covering that circle. The limits of integration will be
1) x going from -2 to 2 and, for each x, y going from -\sqrt{4- x^2} to \sqrt{4- x^2}
or
2) y going from -2 to 2 and, for each y, x going from -\sqrt{4- y^2} to \sqrt{4- y^2}
or, using polar coordinates,
3) r going form 0 to 2 and \theta going from 0 to 2\pi.
 
What's the function that I need to integrate?
 
Cbarker1 said:
What's the function that I need to integrate?

To evaluate volume, $\displaystyle \begin{align*} V = \int{\int{\int_R{1\,\mathrm{d}x}\,\mathrm{d}y}\,\mathrm{d}z} \end{align*}$

HallsofIvy has helped you to find the boundaries, and remember that when converting to cylindrical polars, $\displaystyle \begin{align*} \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \to r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z \end{align*}$...
 
What's about a double integral?
 
Cbarker1 said:
What's about a double integral?

Why not try the triple integral first? If done right, then the right double integral will appear...
 
the instruction says to use a double integral to find the volume of the solid.
 

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