Double Integral Volume Problem

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Homework Help Overview

The problem involves using double integrals to find the volume of a region in the first octant, bounded by the vertical plane defined by the equation 2x + y = 2 and the surface z = x².

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss visualizing the plane and the surface, with some confusion about the nature of the plane's orientation. There are attempts to determine the shape of the base in the xy-plane, with suggestions that it may form a triangle.

Discussion Status

The discussion includes various interpretations of the plane and the surface, with some participants providing insights into the geometry involved. There is acknowledgment of the triangular shape formed by the vertices (1, 0), (0, 0), and (0, 2), but no explicit consensus on the final setup of the integral.

Contextual Notes

Participants express challenges in visualizing the problem and determining the correct bounds for the double integral, indicating a need for clarity in the geometric interpretation of the region.

jumbogala
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Homework Statement


Use double integrals to find the volume of the region in the first octant (x, y, z all more than or equal to zero) bounded by the vertical plane 2x + y = 2 and the surface z = x2


Homework Equations





The Attempt at a Solution


I'm having major problems visualizing this, which is stopping me from even getting started.

z = x2 I think I can visualize by itself.

But the plane is confusing me. My prof taught us that to sketch a plane, you find the zeros of the equation. So setting y and z to zero, we find the plane crosses the x-axis at 1 and similarly the y-axis at 2. But then the plane would be horizontal, not vertical...

Help?
 
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Yes, the plane crosses the x-axis at 1 and the y-axis at 2. It doesn't cross the z axis at all. Doesn't that make it parallel to the z axis? I.e. vertical?
 
Oh, for some reason I was visualizing it passing through the z axis at x = 0. But okay, now I see it.

I think the setup is going to be ∫∫ x 2 dA.

I'm not really sure how the bounds on the integral here work though. Usually I draw a diagram and look at what shape the "base" should have in the xy plane. It kind of looks like it should be a triangle of some kind, but I'm not sure.

Maybe a triangle with vertices (1, 0), (0, 0) and (0, 2)?
 
Last edited:
jumbogala said:
Oh, for some reason I was visualizing it passing through the z axis at x = 0. But okay, now I see it.

I think the steup is going to be ∫∫ x 2 dA.

I'm not really sure how the bounds on the integral here work though. Usually I draw a diagram and look at what shape the "base" should have in the xy plane. It kind of looks like it should be a triangle of some kind, but I'm not sure.

Maybe a triangle with vertices (1, 0), (0, 0) and (0, 2)?

You are definitely right about that triangle. And sure, integrate z (i.e. x^2) over it.
 
I figured out the equations for the sides of the triangle and then did the double integral like normal. It worked. Thanks!
 

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