Triple Integral: Having trouble finding my y bounds

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Homework Help Overview

The problem involves evaluating a triple integral of the form I=\int\int\int_E x^2e^y\,dV, where the region E is defined by a parabolic cylinder and specific planes. The original poster expresses uncertainty about determining the bounds for the variable y.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the geometry of the region defined by the parabolic cylinder and the planes, with one suggesting that the projection onto the xy-plane results in a square region. There is also a request for clarification regarding the notation used in the integral.

Discussion Status

The discussion is ongoing, with participants providing hints and clarifications about the bounds of integration. There is an acknowledgment of the symmetry of the problem and the limits for y being between -1 and 1, but no consensus has been reached on the overall approach to the integral.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the bounds imposed by the parabolic cylinder and the planes, while also addressing potential notation issues in the integral expression.

Saladsamurai
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Homework Statement



[tex]I=\int\int\int_E x^2e^ydV[/tex] where E is bounded by the parabolic cylinder

z=1-y^2 and the planes z=0 x=1 and x=-1

I know that the graph is a parabola that opens downwards and that has symmetry wrt the x-axis. It also stretches along the x-axis toward + and - infinity.

Can I get a hint here? :)
 
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Did you mean
[tex]I=\int\int\int_E x^2e^y\,dV[/tex] ?
 
It's pretty close to being trivial. The parabola z= 1- y2 has z= 0 at y= -1 and y= 1 so projecting down on to xy=plane, we get the square -1< x< 1, -1< y< 1. The z- integral is taken from 0 to 1- y2, the y integral from -1 to 1, and the x integral from -1 to 1.
 
Avodyne said:
Did you mean
[tex]I=\int\int\int_E x^2e^y\,dV[/tex] ?

I don't see any difference. So yes. :smile:
 

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