Double integral to find volume of a solid

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Homework Help Overview

The problem involves setting up a double integral to find the volume of a solid bounded by the graphs y=4-x² and z=4-x², specifically in the context of the first octant.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the double integral and the interpretation of the bounding surfaces. There is uncertainty about the details of the region, particularly regarding the first octant and the implications for the volume calculation.

Discussion Status

Some participants have provided input on the setup of the integral, noting that the original poster's approach appears reasonable. There is acknowledgment of the need for clarity regarding the bounded region, specifically the first octant, which has been confirmed by another participant.

Contextual Notes

Participants are considering the implications of defining the volume in the first octant, where all three variables must be positive. There is a mention of needing more detail about the region to ensure the volume is properly bounded.

mikky05v
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Homework Statement
Set up a double integral to find the volume of the solid bounded by the graphs y=4-x2 and z=4-x2


The attempt at a solution

I drew myself a 3d graph but it's just a parabloid in the xy plane and a parabloid in the xz plane right? so I'm unsure how to set up my integral. This was my attempt, my thought was that perhaps z=4-x2 could be considered like the surface

\int20\int4-x20 (4-x2) dy dx

Could some one give me some input?
 
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mikky05v said:
Homework Statement
Set up a double integral to find the volume of the solid bounded by the graphs y=4-x2 and z=4-x2


The attempt at a solution

I drew myself a 3d graph but it's just a parabloid in the xy plane and a parabloid in the xz plane right? so I'm unsure how to set up my integral. This was my attempt, my thought was that perhaps z=4-x2 could be considered like the surface

\int20\int4-x20 (4-x2) dy dx

Could some one give me some input?

What you have done looks reasonable, but isn't there more detail about the region in question? Like first octant or y positive or something?? Otherwise the volume isn't bounded. Can't tell if your answer is correct without knowing more.
 
oh yes sorry first octant is what it says. I wasn't entirely sure what that meant but I assumed it meant only the positive section of the 3d graph
 
mikky05v said:
oh yes sorry first octant is what it says. I wasn't entirely sure what that meant but I assumed it meant only the positive section of the 3d graph

The first octant is where all three variables are positive. If that's your region your integral is set up correctly.
 
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