# 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

(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.

quantatanu | Student

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.

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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