Using an Ellipse Consider the region bounded by the ellipse

x 2 a 2 + y 2 b 2 = 1

with eccentricity e = c / a .

(a) Show that the solid (oblate spheroid) generated by revolving the region about the minor axis of the ellipse has a volume of V = 4 π 2 b / 3 and a surface area of

S = 2 π a 2 + π ( b 2 e ) ln ( 1 + e 1 − e ) .

(b) Show that the solid (prolate spheroid) generated by revolving the region about the major axis of the ellipse has a volume of V = 4 π a b 2 / 3 and a surface area of

S = 2 π b 2 + 2 π ( a b e ) arcsin e .

To prove: The volume of a solid generated by revolving the region about the minor axis is V=4πa2b3 and surface area for the same solid is S=2πa2+π(b2e)ln(1+e1−e).

Given:

The provided equation of ellipse:

x2a2+y2b2=1

The eccentricity of the ellipse is:

e=ca

Formula used:

The disk integration is given as:

V=π∫x1x2f2(x)dx

The symmetrical property of integration is as given below:

∫−aaf(x)dx=2∫0af(x)dx

The property of integral is as given below:

∫(a+b)dx=∫adx+∫bdx

The formula for eccentricity of an ellipse is:

e=1−b2a2

The integration formula for nth power of secant function is:

∫secnθdθ=n−2n−1∫secn−2θdθ+secn−2θtanθn−1

Proof:

The equation of ellipse is:

x2a2+y2b2=1

Assume b be the minor axis.

Rearrange the ellipse equation as below:

x2a2+y2b2=1x2a2=1−y2b2x2=a2(b2−y2b2) .

x2=a2b2(b2−y2) …...…... (1)

Using disk integration, find the volume:

π∫−bbx2dy=2π∫0ba2b2(b2−y2)dy=2πa2b2∫0b(b2−y2)dy=2πa2b2(∫0bb2dy−∫0by2dy)

Integrate the integrals as follows:

2πa2b2(∫0bb2dy−∫0by2dy)=2πa2b2(b2[y]0b−[y33]0b)=2πa2b2(b2×b−13(b3))=2πa2b2(3b3−b33)=2πa2b2×2b33

On simplifying the above equation:

2πa2b2×2b33=43πa2b

Hence, the volume of the ellipse is 43πa2b.

Take square root on both the sides of equation (1):

x2=a2b2(b2−y2)x=abb2−y2

Differentiate x=abb2−y2 with respect to y:

dxdy=ddy(abb2−y2)=abddy(b2−y2)=ab×12b2−y2(−2y)=−aybb2−y2

Let ds be the small element of the surface of an ellipse, which can be written as:

ds=1+(dxdy)2dy

Substitute the values of dxdy=−aybb2−y2 in the above equation:

1+(dxdy)2dy=1+(−aybb2−y2)2dy=1+a2y2b2(b2−y2)dy=b2(b2−y2)+a2y2b2(b2−y2)dy=b4−b2y2+a2y2bb2−y2dy

Now, the surface area along y−axis is:

s=4π∫0bxds

Substitute the values x=abb2−y2 and ds=b4−b2y2+a2y2bb2−y2dy in the above equation:

s=4π∫0bxds=4π∫0babb2−y2×b4−b2y2+a2y2bb2−y2dy=4π∫0bab2b4−b2y2+a2y2dy=4πab2∫0b(b4−b2y2+a2y2)dy

On simplifying the above equation:

4πab2∫0b(b4−b2y2+a2y2)dy=4πab2∫0b(b4+y2(a2−b2))dy

Simplify the above integral as below:

For ellipse,

Centre of ellipse c2=a2−b2. So,

4πab2∫0b(b4+y2(a2−b2))dy=4πab2∫0b(b4+y2c2)dy …...…..

To prove: The volume of the solid generated by revolving the region about the major axis of the ellipse is V=4πab23 and surface area of the same solid is S=2πb2+2π(abe)sin−1e.