Volumes (1 Viewer)

[cutemouse]

Could someone please help me with this?

Thanks

A solid is formed about the ellipse
x²/25 + y²/9 = 1
in such a way that each plane section of the solid perpendicular to the x-axis is an ellipse whose foci are on the given ellipse. The major and minor axes of each section are proportional to those of the given ellipse. Determine the volume of the solid.
(You may assume the area of the ellipse x2/a2 + y2/b2 = 1 is ab.)

 

[azureus88]

Let major and minor axes of elliptical cross sections be a and b respectively.

[maths]$Since axes are proportional to given ellipse$,b=\frac{3}{5}a\\9=25(1-e^2)\\e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\\ae=y\\a(\frac{4}{5})=y\\a=\frac{5}{4}y\\\delta V=\pi ab\delta x\\=\pi(\frac{3}{5})(\frac{25}{16})y^2\delta x\\=\pi(\frac{3}{5})(\frac{25}{16})(\frac{9}{25})(25-x^2)\delta x\\V=\frac{27}{40}\int_{0}^{5}(25-x^2)dx\\=\frac{225}{4}\pi[/maths]

 

[Affinity]

it is 225pi^2/4

 

[azureus88]

is it? why

 

[shaon0]

is it? why

You forgot a pi outside the front in evaluating V. 3rd last to second last lines.

 

[lolokay]

err the description of the volume sounds like a stretched out sphere to me, so i would think that one pi is correct. azereus only left out pi on the one line, it was back at the end

 

[shaon0]

err the description of the volume sounds like a stretched out sphere to me, so i would think that one pi is correct. azereus only left out pi on the one line, it was back at the end

I thought the last pi came from evaluating the integral

 

[humdiddle]

wrong

 

[Affinity]

err the description of the volume sounds like a stretched out sphere to me, so i would think that one pi is correct. azereus only left out pi on the one line, it was back at the end

The integration generates another pi.. so 1 is definitely missing...