Volume of Solid with Elliptical Base and Isosceles Triangle Cross Sections

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Homework Help Overview

The problem involves finding the volume of a solid with an elliptical base defined by the equation x^2/25 + y^2/16 = 1. The solid has cross sections perpendicular to the x-axis that are isosceles triangles with a fixed altitude of 6 inches. There is also a variation of the problem where the cross sections are taken perpendicular to the y-axis.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for the volume, with one original poster providing a numerical answer and seeking verification. Others question the validity of the answers, suggesting they may be too low and prompting a request for the original work to be shown. There are attempts to manipulate the ellipse equation and derive expressions for the area of the cross sections.

Discussion Status

The discussion is ongoing, with participants providing different interpretations of the problem and questioning each other's calculations. Some guidance is offered regarding the setup of the equations, but there is no explicit consensus on the correct approach or final answer.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can provide to one another. There are indications of confusion regarding the manipulation of the ellipse equation and the calculations for the area of the cross sections.

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Q: A solid has a base in the form of the ellipse: x^2/25 + y^2/16 = 1. Find the volume if every cross section perpendicular to the x-axis is an isosceles triangle whose altitude is 6 inches.
I got 11.781 (or 3.75pi) but I just wanted to check my answer.
Q: Use the same base and cross sections as #3, but change the axis to the y-axis.
Here I got 7.54 (or 2.4pi)
 
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I think your answers are way too low. You'd better show your work.
 
For the first one

[tex] \frac{y^2}{16} = 1-\frac{x^2}{25}[/tex]

[tex] y=\frac{\sqrt{25-x^2}}{20}[/tex]

[tex] A=\frac{3\sqrt{25-x^2}}{10}[/tex]

[tex] V=\int^5_{-5}Adx[/tex]

[tex] V=\frac{3}{10}\left[\frac{1}{2}\sqrt{25-x^2}x + \frac{25}{2}\arcsin\left(\frac{x}{5}\right)\right]\left|^5_{-5}[/tex]

[tex] \approx 11.781[/tex]
 
Making y the subject of the ellipse equation I got:

[tex]y=\pm \frac{4}{5}\sqrt{25-x^2}[/tex]

I can't seem to follow where you went wrong there.
 
y^2/16=1-x^2/25. y=4*sqrt(1-x^2/25). y=(4/5)*sqrt(25-x^2). A=(1/2)*6*(2*y). Stuff like that. You are doing something wrong.
 
EDIT: I accidentally divided by 4 on the other side instead of multiplying.
 
Last edited:

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