How do I calculate the volume of this?

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Discussion Overview

The discussion revolves around calculating the volume of a region bounded by the curves y = x^2, y = 1, and the y-axis when this region is rotated around the y-axis. The scope includes mathematical reasoning and integral calculus.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests defining a volume function V(y) based on the height y, proposing that the derivative dV/dy equals πy, where r = sqrt(y).
  • Another participant expresses a preference for using integrals, initially considering vertical slicing but later questioning how to represent this in integrals.
  • A participant argues against vertical slicing, advocating for horizontal slices that would yield disks with radius x, leading to a volume expression of πy dy.
  • There is a correction regarding the radius of the disks, with a participant questioning whether it should be 1 - sqrt(y) based on the intersection of the curves.
  • Another participant clarifies that every horizontal slice corresponds to a circle with radius sqrt(y) and emphasizes that the area of the disk is πy.

Areas of Agreement / Disagreement

Participants express differing views on the method of slicing the region (vertical vs. horizontal) and the correct representation of the radius in the volume calculation. The discussion remains unresolved with multiple competing approaches presented.

Contextual Notes

There are unresolved assumptions regarding the choice of slicing method and the dependence on the definitions of the curves involved. The discussion includes potential confusion about the intersection points of the curves.

-EquinoX-
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How do I calculate the volume of this??

What is the volume of the region bounded by y = x^2, y = 1, and the y-axis rotated around the y -axis
 
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define a voklume function, as a function of y, by letting V(y) be that portion of the volume lying below height y. then the derivative of this function i the area of the circular face of this portion of volume, i.e. dV/dy = πr^2 where r = x = sqrt(y), so dV/dy = πy. so you guess a formula for V(y) and then plug in y = 1.
 
I should do this using integral.. I was thinking of slicing this region vertically but then how do I represent this in integrals?? I have to take integrals from -1 to 1 right?

I would evaluate this as the integral from 0 to 1 of (1-squareroot of y)^2 dy

is that right??
 
Last edited:
Why in the world would you slice it vertically? Since it is rotated around the y-axis, slices horizontally will be disks with radius x. Each would have area [itex]\pi x^2= \pi y[/itex] and each infinitesmal disc will have volume [itex]\pi y dy[/itex]. Integrate that.
 
HallsofIvy said:
Why in the world would you slice it vertically? Since it is rotated around the y-axis, slices horizontally will be disks with radius x. Each would have area [itex]\pi x^2= \pi y[/itex] and each infinitesmal disc will have volume [itex]\pi y dy[/itex]. Integrate that.


I am sorry that's my mistake. I would slice it horizontally and take the integral from 0 to 1. And shouldn't the radius be 1-square root of y?? Because it's the intersection of y = x^2 and y = x
 
-EquinoX- said:
I am sorry that's my mistake. I would slice it horizontally and take the integral from 0 to 1. And shouldn't the radius be 1-square root of y?? Because it's the intersection of y = x^2 and y = x
Where did you get y= x from? Your original post was:

-EquinoX- said:
What is the volume of the region bounded by y = x^2, y = 1, and the y-axis rotated around the y -axis
Every horizontal "slice" is a circle with center at the y-axis, x= 0, and the end of a radius at [itex]x= \sqrt{y}[/itex]. Of course, the area of the disk is [itex]\pi x^2= \pi y[/itex].
 

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