How to find the volume under a surface?

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Homework Help Overview

The discussion revolves around understanding the representation of the surface defined by the equation z = $$\sqrt{1-x^2}$$ and its implications in three-dimensional space. Participants explore how this equation relates to volume and the geometry of surfaces, particularly in relation to the x-y and x-z planes.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster questions how the equation can represent a surface given its apparent dimensional constraints and expresses confusion about the role of the y-coordinate. Other participants clarify that the equation describes a two-dimensional surface in three-dimensional space and explore the implications of this representation.

Discussion Status

Participants are actively engaging with the concepts, with some providing clarifications about the nature of the surface and its dimensionality. There is a recognition of the misunderstanding regarding the y-coordinate and its implications for the surface representation.

Contextual Notes

There is a focus on the geometric interpretation of the equation and its implications for volume, with some participants addressing assumptions about the coordinates and their roles in defining the surface.

Adesh
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Homework Statement
Find the volume under the surface z = \sqrt{1-x^2} and above the triangle formed y = x , x=1 , and the x-axis.
Relevant Equations
Volume = double integral of the surface function.
I want to know that how can z=$$ \sqrt{1-x^2}$$ ever represent a surface? It graphs a curve in the x-z plane and the triangle lies in x-y plane so how can they contain a volume, they are orthogonal to each other. I have attached awn image which is drawn GeoGebra for the function z=$$\sqrt{1-x^2}$$. My question is if we can write the equation z = $$\sqrt{1-x^2}$$ as $$\sqrt{1-x^2} - 0.y - z $$. Then how can it traverse in the y-direction, it's coordinate always have to be zero.
Any help would be appreciated. Thank you.
 

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You are making a large error when you say "it's coordinate always have to be zero". No, 0*y= 0 for any value of y- that does not mean y= 0!

First, we are talking about three dimensions so "z= \sqrt{1- x^2}" means the set of points (x, y, z)= (x, y, \sqrt{1- x^2}). There are two "parameters", x and y, so this is a two dimensional surface.

You should be able to see that z= \sqrt{1- x^2} can be written z^2= 1- x^2 or x^2+ z^2= 1. That is a circle in the xz-plane. Since y can be anything (NOT only 0) that extends to the cylinder with axis along the y-axis. Since we are taking the positive square root the figure is the upper half of the cylinder.
 
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HallsofIvy said:
You are making a large error when you say "it's coordinate always have to be zero". No, 0*y= 0 for any value of y- that does not mean y= 0!

First, we are talking about three dimensions so "z= \sqrt{1- x^2}" means the set of points (x, y, z)= (x, y, \sqrt{1- x^2}). There are two "parameters", x and y, so this is a two dimensional surface.

You should be able to see that z= \sqrt{1- x^2} can be written z^2= 1- x^2 or x^2+ z^2= 1. That is a circle in the xz-plane. Since y can be anything (NOT only 0) that extends to the cylinder with axis along the y-axis. Since we are taking the positive square root the figure is the upper half of the cylinder.
Thank you so much. You have made a very nice remark, I have understood, just x= 0 doesn’t represent only one point but whole of y-axis.
Thank you so much.
 
Adesh said:
You have made a very nice remark, I have understood, just x= 0 doesn’t represent only one point but whole of y-axis.
No. The equation x = 0 represents the whole y-z plane in ##\mathbb R^3##, not just the y-axis.
 
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