# An ellipse centered at the origin is described by the equation ((x^2)/(a^2))+ ((y^2)/(b^2)) =1Please answer letter c. The other letters have been answered already on enotes lately. (a) Set up an...

An ellipse centered at the origin is described by the equation ((x^2)/(a^2))+ ((y^2)/(b^2)) =1

Please answer letter c. The other letters have been answered already on enotes lately.

(a) Set up an integral for the volume of the ellipsoid generated when the region defined by ((x^2)/(a^2))+ ((y^2)/(b^2)) =1 is revolved about the x-axis. Do not evaluate the integral.

b) Set up an integral for the volume of the ellipsoid generated when the region de ned by ((x^2)/(a^2))+ ((y^2)/(b^2)) =1 is revolved about the y-axis. Do not evaluate the integral.

(c) Should the results of (a) and (b) agree? Explain.

Asked on by goldbergfan

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

if we interchange "a" by "b" and "x" by "y" then the ellipse looks just the same, so

a)  If the ellipse is revolved around x-axis, we must replace "a" by "b" just in the earlier answer that I am pasting at the end of this answer.

so

volume of ellipsoid of revolution = (2 Pi Integral [dy' (1-y'^2)] from

0 to 1) * ba  (instead of "ab" but

its the same)

b)  If the ellipse is revolved around "y-axis", then the problem we have already solved earlier, that is

volume of ellipsoid of revolution = (2 Pi Integral [dy' (1-y'^2)] from

0 to 1) * ab

c) the results of (a) and (b) should agree because the reason I have already told in the beginning, I repeat:

if we interchange "a" by "b" and "x" by "y" then the ellipse looks just the same.

_________________________________

__________________________________

let us take,

x/a = x'

y/b = y'

hence dx = a dx'

dy = b dy'

Now the given equation of ellipse becomes:

x'^2 + y'^2 = 1

which is the equation of a circle of radius 1 centered at the origin (0,0).

So if we revolve it around y-axis which is equivalent to revolving around y'-axis. as y is just a constant b-times y', we get a sphere.

To set the integral we can do one thing:

we take a slice that is a disc perpendicular to y'-axis of thickness dy' and then add up all these disc-volume to get the whole sphere volume, but we can infact calculate half of the volume and then double it. So we take a disc-slice of thickness dy' of radius x', and this x' ofcourse obey,

1=x'^2 + y'^2, as the radius of the sphere is 1. So x'^2 = 1- y'^2

so

volume of sphere = 2 Integral [dy' (Pi x'^2 )] from 0 to 1

= 2 Pi Integral [dy' (1-y'^2)] from  0 to 1

then after doing the integral we must scale back to original ellipse problem so we must do the following replacements:

1--> a along x

1--> b along y

so finally:

volume of ellipsoid of revolution = (2 Pi Integral [dy' (1-y'^2)] from

0 to 1) * ab

answer will be (4/3)Pi*a*b

We’ve answered 317,686 questions. We can answer yours, too.