Volume of Solid with Elliptical Base and Isosceles Triangle Cross Sections

[mrdoe]

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)

 

[Dick]

I think your answers are way too low. You'd better show your work.

 

[mrdoe]

For the first one

$$\frac{y^2}{16} = 1-\frac{x^2}{25}$$

$$y=\frac{\sqrt{25-x^2}}{20}$$

$$A=\frac{3\sqrt{25-x^2}}{10}$$

$$V=\int^5_{-5}Adx$$

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

$$\approx 11.781$$

 

[Mentallic]

Making y the subject of the ellipse equation I got:

$$y=\pm \frac{4}{5}\sqrt{25-x^2}$$

I can't seem to follow where you went wrong there.

 

[Dick]

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.

 

[mrdoe]

EDIT: I accidentally divided by 4 on the other side instead of multiplying.