Question

Use an iterated integral to compute the area of the ellipse x 2 a 2 + y 2 b 2 = 1. The a and b are positive constants. (Hint: You will need to use a trigonometric substitution to do one of the integrations. You may need to review that method in the text. There are going to be quite a few complicated integrals for the rest of the course, so you’ll probably get many opportunities to review integration techniques covered in Calculus II. Might as well put a bookmark in that section of the text right now.)

Answer 1

In rectangular coordinates, the area is

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx`

or

`\displaystyle\int_(-b)^b\int_(-(a/b)√(b^2-y^2))^((a/b)√(b^2-y^2))\mathrm dx\,\mathrm dy`

We'll compute the first integral. By symmetry of the integrand, we have

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=4\int_0^a\int_0^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx`

The integral wrt
`y` is trivial, giving

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=\frac{4b}a\int_0^a√(a^2-x^2)\,\mathrm dx`

In the remaining integral wrt
`x`, substitute

`x=a\sin t\implies\mathrm dx=a\cos t\,\mathrm dt`

to get

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=4b\int_0^(\pi/2)\cos t√(a^2-(a\sin t)^2)\,\mathrm dt`

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=4ab\int_0^(\pi/2)\cos t√(1-\sin^2t)\,\mathrm dt`

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=4ab\int_0^(\pi/2)\cos t√(\cos^2t)\,\mathrm dt`

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=4ab\int_0^(\pi/2)\cos t\,|\cos t|\,\mathrm dt`

Over the interval
`0<t<\frac\pi2`, we have
`\cos t>0`, so that
`|\cos t|=\cos t`. Then
`\cos t\,|\cos t|=\cos^2t`, and recall a double angle identity, we get rewrite the integral as

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=2ab\int_0^(\pi/2)(1+\cos2t)\,\mathrm dt`

The remaining integral is trivial.

`\displaystyle\int_0^(\pi/2)(1+\cos2t)\,\mathrm dt=\left(t+\frac12\sin2t\right)\bigg|_(t=0)^(t=\pi/2)=\frac\pi2`

Then the area of the ellipse is

`\displaystyle\int_(-a)^a\int_(-(b/a)√(a^2-x^2))^((b/a)√(a^2-x^2))\mathrm dy\,\mathrm dx=\boxed{\pi ab}`