How to find the volume under a surface?

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The discussion clarifies how the equation z = √(1-x²) represents a three-dimensional surface rather than just a curve in the x-z plane. It emphasizes that the surface extends along the y-axis, allowing for a two-dimensional representation with x and y as parameters. The equation can be reformulated to show that it describes a circle in the x-z plane, which extends into a cylinder in three dimensions. The participants correct misconceptions about the coordinates, highlighting that setting y to zero does not restrict it to a single point but allows for a full range along the y-axis. Overall, the conversation enhances understanding of the geometric implications of the equation in three-dimensional space.
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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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