How to find the volume under a surface?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
Adesh
Messages
735
Reaction score
191
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.
 

Attachments

  • Screen Shot 2019-08-14 at 9.58.26 PM.png
    Screen Shot 2019-08-14 at 9.58.26 PM.png
    40.8 KB · Views: 337
on Phys.org
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 "[itex]z= \sqrt{1- x^2}[/itex]" means the set of points [itex](x, y, z)= (x, y, \sqrt{1- x^2})[/itex]. There are two "parameters", x and y, so this is a two dimensional surface.

You should be able to see that [itex]z= \sqrt{1- x^2}[/itex] can be written [itex]z^2= 1- x^2[/itex] or [itex]x^2+ z^2= 1[/itex]. 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.
 
  • Like
Likes   Reactions: Adesh
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 "[itex]z= \sqrt{1- x^2}[/itex]" means the set of points [itex](x, y, z)= (x, y, \sqrt{1- x^2})[/itex]. There are two "parameters", x and y, so this is a two dimensional surface.

You should be able to see that [itex]z= \sqrt{1- x^2}[/itex] can be written [itex]z^2= 1- x^2[/itex] or [itex]x^2+ z^2= 1[/itex]. 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 [itex]x= 0[/itex] 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 [itex]x= 0[/itex] 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.
 
  • Like
Likes   Reactions: Adesh