The ellipse can be divided into two equal halves because it is centred at the origin. If we rotate a quadrant of the ellipse we'll get one half of the volume of the spheroid. If we take a point (x,y) on the ellipse then we can imagine a vertical disk with radius y, infinitesimal thickness dx, at a distance x from the origin. The volume of the disk is πy2dx. This is the integrand we need to find the volume of half of the spheroid. The limits of the definite integral are x=0 to x=a.
Volume=∫0aπy2dx. y2=b2(1-x2/a2) so we have:
(b2/a2)π∫0a(a2-x2)dx=(b2/a2)π[a2x-x3/3]0a=(b2/a2)π(⅔a3)=2πab2/3.
This is half the volume of the spheroid so the volume of the latter is 4πab2/3. Note that when a=b we get the volume of a sphere. There is another solution, because we could have rotated the quadrant around the y-axis making the volume 4πa2b/3. There are two possible spheroids (with different volumes) for a≠b and a,b≠0.