Volumes by Slicing and Rotation About an Axis

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Homework Help Overview

The discussion revolves around calculating the volume of a solid whose base is defined by the disk X^2 + Y^2 <= 1, with cross-sections that are isosceles right triangles perpendicular to the Y-axis. Participants express confusion regarding the setup and the appropriate methods to apply.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the volume by calculating the area of the triangular cross-sections and integrating over the specified range. Some participants suggest clarifying the orientation of the cross-sections and how to express dimensions in terms of the correct variables.

Discussion Status

Participants are actively engaging with the problem, with some providing corrections and suggestions for reformulating the volume element. There is a recognition of the need to express the volume in terms of the correct variables, and multiple interpretations of the setup are being explored.

Contextual Notes

There is uncertainty regarding the application of the disk and washer methods, as well as the correct interpretation of the cross-section dimensions. The original poster expresses a lack of confidence in their understanding of the problem.

TheTaoOfBill
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Homework Statement


I'm having issues with this section in calc. I'm not at all sure what I'm doing!

Here is the problem I'm having trouble with:

Directions: Find the volume of the solids.

Problem: The base of the solid is the disk X^2 + Y^2 <= 1. The cross-sections by planes perpendicular to the Y-axis between Y= -1 and Y= 1 are isosceles right triangles with one leg in the disk.

So essentially it shows a right isosceles triangle where right angle touches the left edge of the circle and the base of the right triangle extends to the other side of the circle.



Homework Equations



about all I really understand what to do is find the area of the triangle.

A(x) = 1/2 BH
B=2x=H
1/2(2x)(2x) = 2x^2

The Attempt at a Solution



I honestly don't have a clue what I'm doing in this section. But if I were to guess I'd take the integral of the area from 1to-1 since that's my interpretation of what the book says.

So V = int from 1to-1(2x^2) = 4/3

But that doesn't sound right at all.

It also talks about the disk and washer methods in this section but I'm not sure if those apply. Any help would be appreciated.
 
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You have the idea, but the plane cross sections are perpendicular to the y axis, not parallel to it. So the base of the triangle is 2y, which you can get in terms of x from the equation of the circle. Your cross section slab has thickness dx. So your element of volume is

dV = (1/2)(2y)2 dx

Get it all in terms of x and go.
 
LCKurtz said:
You have the idea, but the plane cross sections are perpendicular to the y axis, not parallel to it. So the base of the triangle is 2y, which you can get in terms of x from the equation of the circle. Your cross section slab has thickness dx. So your element of volume is

dV = (1/2)(2y)2 dx

Get it all in terms of x and go.

ok so
dV = 1/2(2(sqrt(1-x^2))

so V = int from -1 to 1(1/2(2(sqrt(1-x^2))) = pi/2 ?
 
Did you miss a squared term?
 
LCKurtz said:
Did you miss a squared term?

Ah.. right...

So..

V = int from -1 to 1(1/2(2(sqrt(1-x^2))^2) = 8/3
 

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