$\int_{0}^{\pi/2} \int_{0}^{2a \cos \theta} f(r, \theta) dr d \theta \equiv \int_{\theta = \alpha}^{\theta = \beta} \int_{r = f_{1}(\theta)}^{r = f_{2} (\theta)} f(r, \theta) dr d \theta$

The bounding curves are given by

$r = 0, \ r = 2a \cos \theta$

$\theta = 0, \ \theta = \pi/2$

Let us plot these curves to get the region of integration

After changing the order, the integral will be of the form

$I = \int_{r = r_{1}}^{r = r_{2}} \int_{\theta = g_{1}(r)}^{\theta = g_{2} (r)} f(r, \theta) d \theta dr$

To get the limit of $\theta,$ let us draw a circular arc with centre at the origin oriented in anticlockwise direction.

The arc intersects first $\theta = 0$ and then intersect $\theta = \cos^{-1} \dfrac{r}{2a}$ . So, limit of $\theta$ is from $\theta = 0$ to $\theta = \cos^{-1} \dfrac{r}{2a}$ . To find out limit of r, we must find out in what range the radius of arc should vary so that whole region of integration is transerved. r should vary from r = 0 to r = 2a.

So, after changing the order, the integral changes to

$I = \int_{0}^{2a} \int_{0}^{cos^{-1} \dfrac{r}{2a}} f(r, \theta) d \theta dr$

Example 2 :

Change the order of integration in the system of integrals

$\int_{0}^{\pi/2} \int_{a \cos \theta}^{a(1 + \cos \theta)} f(r, \theta) dr d\theta + \int_{\pi/2}^{\pi} \int_{0}^{a(1 + \cos \theta)} f(r, \theta)dr d \theta$

Solution :

Comparing the integral with its standard form

$\int_{\theta = \alpha}^{\theta = \beta} \int_{r = f_{1} (\theta)}^{r = f_{2}(\theta)} f(r, \theta) dr d \theta$

The region of integration is bounded by curves

$r = a \cos \theta \ \ \ \ \theta = 0$

$r = a(1 + \cos \theta), \ \ \ \ \theta = \pi$

After changing the order the integral is of the form

$I = \int_{r = r_{1}}^{r = r_{2}} \int_{\theta = f_{1} (r) }^{\theta = f_{2}(r)} f(r, \theta) d \theta dr$

To find the limit of $\theta$ , let us draw an circular arc of radius r intersecting the region of integration arbitrary. The arc has two possibilities. If radius of arc is less than a it intersects circle and cardiod $(r = a(1 + \cos \theta))$ and if a < r < 2a, it intersects initial line and cardiod. So, we have to partition the domain of integration by a partitioning curve in the form of circular arc of radius a into region I and II.

For region, I let us draw a circular arc intersecting the region I arbitrary. The limit of $\theta = \cos^{-1} \dfrac{r}{a}$ to $\theta = \cos^{-1} \dfrac{r - a}{a}$ and r will vary from r = 0 to r = a.

For region II, let us draw a circular arc intersection the region II arbitrary. The limit of $\theta$ is from $\theta = 0$ to $\theta = \cos^{-1} \frac{r - a}{a}$ and r will vary from r = a to r = 2a.

So after changing the order, the integral will convert to

$I = \int_{0}^{a} \int_{\cos^{-1} \frac{r}{a}}^{\cos^{-1} \frac{r - a}{a}} f(r, \theta) d \theta dr + \int_{a}^{2a} \int_{0}^{\cos^{-1}\frac{r - a}{a}} f(r, \theta) d \theta dr$

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