Double integral over area bounded by ellipse

Question

Evaluate the double integral Area enclosed by ellipse

Answer 1

Let,

$$x=ar\cos(\theta)$$

$$y=br\sin(\theta)$$

With $\theta \in [0,2\pi]$ and $r \in [0,1]$. Then the Jacobian corresponding to this transformation is $abr$. And we have the double integral,

$$\int_{0}^{2\pi} \int_{0}^{1} (ar\cos(\theta)+br\sin(\theta))^2 ab r dr d\theta$$

$$=ab \int_{0}^{2 \pi} \int_{0}^{1} \left(a^2r^2\cos^2(\theta)+b^2r^2 \sin (\theta)+\frac{ab}{2}r^2 \sin(2\theta) \right) r dr d\theta$$

Noticing that cosine is $2\pi$ periodic so we quickly see we can drop the $\sin(2\theta)$ term then separate the double integral as follows.

$$=ab \int_{0}^{1} r^3 dr \int_{0}^{2\pi} \left(a^2\cos^2(\theta)+b^2\sin^2(\theta) \right) d\theta$$

$$=\frac{ab}{4} \int_{0}^{2\pi} \left(\frac{1}{2}a^2(1+\cos(2\theta)+\frac{1}{2}b^2(1-\cos(2\theta) \right) d\theta$$

By $2\pi$ periodicity of sine we will see we can drop the $\cos(2\theta)$ terms to get,

$$=\frac{ab}{4} \int_{0}^{2\pi} \left(\frac{a^2}{2}+\frac{b^2}{2} \right) d\theta$$

$$=\frac{\pi ab}{4}(a^2+b^2)$$

Answer 2

If we are going to do this without using a transformation. Notice our regions boundary is,

$$(\frac{x}{a})^2+(\frac{y}{b})^2=1$$

So that $x \in [-a\sqrt{1-(\frac{y}{b})^2},a\sqrt{1-(\frac{y}{b})^2}]$ and of course $y \in [-b,b]$. So we are interested in computing,

$$\int_{-b}^{b} \int_{-a\sqrt{1-(\frac{y}{b})^2}}^{a\sqrt{1-(\frac{y}{b})^2}} (x^2+2xy+y^2) dx dy$$

We integrate with respect to $x$ first. Notice that $2xy$ is odd with respect to $x$ so we can drop that term. So then what we get is,

$$=\int_{-b}^{b} \left( \frac{2}{3}(a\sqrt{1-(\frac{y}{b})^2})^3+y^2\left(2a\sqrt{1-(\frac{y}{b})^2}\right) \right) dy$$

$$=\int_{-b}^{b} \left(\frac{2}{3}(a\sqrt{1-(\frac{y}{b})^2})a^2\left(1-(\frac{y}{b})^2\right)+y^2\left(2a\sqrt{1-(\frac{y}{b})^2}\right) \right) dy$$

First let $\frac{y}{b}=u$ so $dy=bdu$ and we have that what we want is the sum of,

$$I_1=\frac{2}{3}a^3b \int_{-1}^{1}(1-u^2)\sqrt{1-u^2} du$$

$$I_2=2ab^3\int_{-1}^{1} u^2\sqrt{1-u^2} du$$

The integrals in each of these are both standard, and related integrals. Find one, you find the other by geometry.

Answer 3

\begin{align} I= \iint_A (x+y)^{2} dxdy = \iint_A (x^2 +y^2) dxdy + 2\iint_A xy\ dxdy \end{align}

The latter integral vanishes due to odd symmetry in $x$ and $y$. We didn't need odd symmetry in both for the integral to vanish; just one would have sufficed. \begin{align} \iint_A xy\ dxdy = \int_{-b}^b\int_{-a\sqrt{1-(y/b)^2)}}^{a\sqrt{1-(y/b)^2)}} xy\ dxdy =0 \end{align}

The former integral is best evaluated in polar, but we proceed in Cartesian. It turns into a trigonometric integral shortly. First, note the even symmetry in both $x$ and $y$, leading to simplified bounds and a factor of $2^2$ in front. \begin{align} I= \iint_A (x^2 +y^2) dxdy &= \int_{-b}^b\int_{-a\sqrt{1-(y/b)^2)}}^{a\sqrt{1-(y/b)^2)}} (x^2 +y^2) dxdy \\ &= 4\int_0^b\int_0^{a\sqrt{1-(y/b)^2)}} (x^2 +y^2) dxdy \\ &= 4\int_0^b \left( \frac13a^3\left(1-\frac{y^2}{b^2}\right)^{3/2} +ay^2\sqrt{1-\frac{y^2}{b^2}} \right)dy \end{align}

After reducing to a single integral, we need to make a trig substitution. The trigonometric integrals are then evaluated through repeated use of the half angle formula.

\begin{align} I &= 4\int_0^{\pi/2} \left( \frac{a^3b}{3}\cos^4t +ab^3\sin^2 t\cos^2 t \right) dt \qquad\text{(letting } \sin t=y/b) \\ &= 4\int_0^{\pi/2} \left( \frac{a^3b}{3}\cos^4t +\frac{ab^3}{4}\sin^2(2t) \right) dt \qquad\text{(double angle formula)} \\ &= 4\left( \frac{a^3b}{3}\frac{3\pi}{16} +\frac{ab^3}{4}\frac{\pi}{4}\right) \qquad\text{(repeated use of half angle formulae)} \\ &= \frac{ab\pi}{4}\left(a^2 +b^2\right) \end{align}