# An ellipse centered at the origin is described by the equation ((x^2)/(a^2))+ ((y^2)/(b^2)) =1Set up an integral for the volume of the ellipsoid generated when the region defined by ((x^2)/(a^2))+...

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

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.

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### 1 Answer

You should evaluate the volume of ellipsoid using disk method, hence, you need to compute the volume of each disk slice and then you need to integrate such that:

`dV = pi*r^2*dx`

The radius of disk is considered to be y and the thickness is dx, such that:

`dV = pi*y^2*dx`

You need to use the equation of ellipse such that:

`x^2/a^2 + y^2/b^2 = 1 => y^2/b^2 = 1 - x^2/a^2`

`y^2 = b^2 - x^2*(b^2/a^2)`

You may evaluate the volum of ellipsoid such that:

`V = int_(-b/a)^(b/a) pi*y^2 dx`

`V = int_(-b/a)^(b/a) pi*(b^2 - x^2*(b^2/a^2))dx`

Since `y^2` is an even function, hence `int_(-b/a)^(b/a) pi*(b^2 - x^2*(b^2/a^2))dx = 2pi*int_0^(b/a) (b^2 - x^2*(b^2/a^2))dx`

`V = 2pi*(int_0^(b/a) b^2 dx- int_0^(b/a) x^2*(b^2/a^2)dx) ` `V = 2pi*(b^2x - (b^2/a^2)x^3/3)|_0^(b/a)`

**Hence, setting up the volume of ellipsoid generated by revolving the ellipse of equation `x^2/a^2 + y^2/b^2 = 1 ` about x axis, yields `V = 2pi*int_0^(b/a) (b^2 - x^2*(b^2/a^2))dx = 2pi*(b^2x - (b^2/a^2)x^3/3)|_0^(b/a).` **

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