Find the volume of the spheroid bounded by the ellipse x^2/(a^2)+ y^2/(b^2=1

Question

please help

Answer 1

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=∫₀^aπy²dx. y²=b²(1-x²/a²) so we have:

(b²/a²)π∫₀^a(a²-x²)dx=(b²/a²)π[a²x-x³/3]₀^a=(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.