Question

length of curve 8a^2y^2=x^2(a^2-x^2)?​

Answer 1

Answer:

When x = 0, y = 0 or y = 2a. Also y > 2a is impossible because x^2 > 0.

The graph is symmetric about the x axis so you only need to double the area between the y axis and the right hand half of the curve.

Write as x = y^(3/2) sqrt(2a-y)/a. You just have to integrate this from 0 to 2a, assuming a > 0. If there is no easier way, expand 2a-y as a power series in y/2a.

Alternatively, put u = y/2a, du = dy/2a, then integrate

2a sqrt(2/a) (2a)^(3/2) u^(5/2 - 1) (1-u)^(3/2 - 1) du from 0 to 1.

This is a beta function, 2a sqrt(2/a) (2a)^(3/2) beta(5/2, 3/2)

= 4a^2 gamma(5/2)gamma(3/2)/gamma(8/2)

= 4a^2 (3/2)(1/2) sqrt(pi) (1/2) sqrt(pi)/3!

Doubling, the whole area is a^2 pi/2.

Answer 2

Step-by-step explanation:

8a ^ 2 * y ^ 2 = x ^ 2 * (a ^ 2 - x ^ 2)