Question

How to calculate the area of an ellipse by using integration?

Answer 1

An ellipse has the general equation

(x/a)² + (y/b)² = 1

(actually, this is an ellipse whose semimajor and -minor axes are parallel to the x- and y-axes, but one can rotate any ellipse so that it has the same configuration, and the area remains constant)

Convert to a kind of rescaled polar coordinates, using

x = a r cos(t)

y = b r sin(t)

Then the interior of such an ellipse is the set

E = {(r, t) : 0 ≤ r ≤ 1 and 0 ≤ t ≤ 2π}

The area of E is then given by the integral

`\displaystyle \iint_E dA`

Compute the Jacobian determinant for this change of coordinates. The Jacobian is

`J = \begin{bmatrix}x_r & x_t \\ y_r & y_t\end{bmatrix} = \begin{bmatrix}a \cos(t) & -a r \sin(t) \\ b \sin(t) & b r \cos(t)\end{bmatrix}`

Then we have Jacobian determinant

|det(J)| = |a b r cos²(t) + a b r sin²(t)| = |a b r| = |a b| r

since r ≥ 0.

The area of E is then

`\displaystyle \iint_E dA = \iint_E |ab| r \, dr \, dt`

`\displaystyle \iint_E dA = \int_0^(2\pi) \int_0^1 |ab| r \, dr \, dt`

`\displaystyle \iint_E dA = 2\pi |ab| \int_0^1 r \, dr`

`\displaystyle \iint_E dA = \boxedab`