Double integral volume of a region bounded by two z planes

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Homework Help Overview

The discussion revolves around finding the volume of a region bounded by the surfaces defined by the equations z = 1 - y^2 and z = y^2 - 1, with constraints on x between 0 and 2. Participants are exploring how to set up the double integral for this volume calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are considering whether to split the volume into two integrals corresponding to each z function and then combine them. There is discussion about determining the appropriate bounds for integration and the specific region of interest within the defined surfaces.

Discussion Status

Some participants have suggested sketching the surfaces to better understand the volume and the limits of integration. There is an ongoing exploration of the correct interpretation of the inequalities defining the region, with no explicit consensus reached on the bounds or the setup of the integrals.

Contextual Notes

Participants are questioning the range for y and how it relates to the chosen region between the two surfaces. There is a need for clarity on whether to consider the region where 1 - y^2 is less than or greater than y^2 - 1.

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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