Daniel B. answered • 05/16/21
A retired computer professional to teach math, physics
(I)
Consider the range of values where the curve is defined,
under the assumption that we do not deal with complex numbers.
x >=0 and y >=0 to make the left hand side real.
x <= a and y <= a because otherwise one of (x/a)2/3 or (y/a)2/3
would have to be negative.
Thus both x and y must be in the range [0, a].
(II)
For simplicity let me rewrite the given equation as
x2/3+ y2/3 = b
where
b = a2/3
(III)
Let's calculate dx/dy.
x = (b - y2/3)3/2
dx/dy = (3/2)(b - y2/3)1/2(-2/3)y-1/3
= -(b - y2/3)1/2y-1/3
(IV)
Imagine the y-range as divided into small intervals of width dy.
In the limit we can assume the intervals to be sufficiently small so that
within each interval the curve is a line segment.
Each of these line segments at position y is the hypotenuse of a right triangle with sides
dy and
dx = (dx/dy)dy = -(b - y2/3)1/2y-1/3dy
By Pythagorean theorem the length of the hypotenuse is
√(dx² + dy²) =
√((b - y2/3)y-2/3dy² + dy²)) =
√((b - y2/3)y-2/3 + 1) dy =
√(y-2/3b - 1 + 1) dy =
y-1/3b dy
(V)
Each of those line segments calculated above is being rotated around the x-axis,
which makes the radius of rotation equal y.
Therefore each line segment contributes an area of
2πy(y-1/3b)dy = 2πy2/3b dy
(VI)
The total surface of the resulting region is then the integral from 0 to a
∫2πy2/3b dy =
2πb∫y2/3 dy =
2πb(3/5)y5/3 evaluated between 0 and a =
2πb(3/5)a5/3 =
(6/5)πa7/3 (after having substituted b = a2/3)
