Double integral volume of a region bounded by two z planes

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SUMMARY

The discussion focuses on calculating the volume of the region bounded by the surfaces defined by the equations z = 1 - y² and z = y² - 1, specifically for x in the range [0, 2]. Participants suggest splitting the volume into two integrals corresponding to each z function and then combining the results. The recommended approach is to integrate the difference of the two functions, g(x,y) - f(x,y), and to determine the appropriate bounds for y based on the chosen region. A sketch of the surfaces is advised to visualize the finite region of integration.

PREREQUISITES
  • Understanding of double integrals in multivariable calculus
  • Familiarity with the concepts of bounded regions in three-dimensional space
  • Knowledge of sketching surfaces defined by equations
  • Ability to manipulate algebraic expressions involving square roots
NEXT STEPS
  • Learn how to set up double integrals for volume calculations in multivariable calculus
  • Study the method of finding bounds for integrals in three-dimensional regions
  • Explore the concept of sketching surfaces to visualize integration limits
  • Investigate the properties of functions defined by z = f(x,y) and z = g(x,y) for volume calculations
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Students and educators in calculus, mathematicians working with multivariable functions, and anyone involved in geometric volume calculations using double integrals.

mahrap
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a) find the volume of the region enclosed by
z = 1 - y^2 and z = y^2 -1 for x greater or equal to 0 and less than or equal to 2.

b) would i split up the volume into two integrals, each integral for each z function and then add them together? I also don't know how to find the bounds. Please help with thorough steps
 
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mahrap said:
b) would i split up the volume into two integrals, each integral for each z function and then add them together?
The simplest is to note that you can split it into two equal volumes (z > 0, z < 0), so you only have to calculate one. But in general, given z > f(x,y) and z < g(x,y), it's as easy to integrate g(x,y)-f(x,y).
Surfaces like this carve the space into several regions, some of which are infinite. You have to decide which (finite) region is being asked for. Best is to sketch it. In the present case, you have to choose between 1 - y2 < z < y2-1 and 1 - y2 > z > y2-1. Which do you think it is?
 
haruspex said:
The simplest is to note that you can split it into two equal volumes (z > 0, z < 0), so you only have to calculate one. But in general, given z > f(x,y) and z < g(x,y), it's as easy to integrate g(x,y)-f(x,y).
Surfaces like this carve the space into several regions, some of which are infinite. You have to decide which (finite) region is being asked for. Best is to sketch it. In the present case, you have to choose between 1 - y2 < z < y2-1 and 1 - y2 > z > y2-1. Which do you think it is?




When i sketch the surface it seems to me the limits of integration should be 0 to 2 for x and
(z+1)^(1/2) to (1-z)^(1/2) for y. would my function inside the integral be
( 1 - y^2 - (y^2 - 1) )
 
mahrap said:
When i sketch the surface it seems to me the limits of integration should be 0 to 2 for x and
(z+1)^(1/2) to (1-z)^(1/2) for y. would my function inside the integral be
( 1 - y^2 - (y^2 - 1) )
You need a range for y which does not involve z.
Pls answer my question: do you think you want the region 1 - y2 < z < y2-1 or 1 - y2 > z > y2-1? Once you've chosen that you'll see what the range for y is.
 

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