Question

Sketch the region of integration and convert the polar integral into cartesian integral or sum of cartesian integral. Don't evaluate: integral^pi/6_0 integral^2sec theta_0 r^5sin 2 theta drd theta

Answer 1

Answer: The transformed integral is

$\int_0^{\pi/6}\int_0^{2\sec\theta}r^5\sin^2\theta dr d\theta=\int_0^2\int_0^{\frac{1}{\sqrt{4}}x}y^2(x^2+y^2).$

The region of integration is a triangle delimited by the $x$ axis, the vertical line $x=2$, and the line $y=\frac{1}{\sqrt{3}}x$.

Step-by-step explanation: For this integral we have:

1. $0$ and

2. $0$.

The first condition says that the region of integrarion is within the "slice" of the xy plane between the polar angle of $\theta=0$ and $\theta=\frac{\pi}{6}$ in the first quadrant. To make sense of the second condition we will convert to the Cartesian coordinates. Remember that

3. $x=r\cos\theta$

4. $y=r\sin\theta$

5. $r=\sqrt{x^2+y^2}$

6. $\theta=\arctan\frac{y}{x}\quad \text{only in the first quadrant}$

Using 2. (squaring it), along with 5. we obtain

$0$

Now using 3. to express $\cos^2\theta=\frac{x^2}{r^2}=\frac{x^2}{x^2+y^2}$ we obtain

$0$

The left inequality $0$ just says that we start from the origin. The right inequality, after cancelling $x^2+y^2$, yields $1$. Since we are in the first quadrant $x$ has to be positive so this inequality can be rewritten as (taking the square root) $x$. So the region is a triangle delimited by the slice of the first quadrant between $\theta=0$ and $\theta=\frac{\pi}{6}$ and by the vertical line $x=2$ (see figure).

In other words, this triangle is delimited by the x axis, by the vertical line x=2 and by the line $\frac{y}=\tan\frac{\pi}{6}x=\frac{1}{\sqrt{3}}x.$ This helps us to define the boundaries of integration: we have $0$ (from the origin up to x=2) and $0$ (from the x axis i.e. y=0 up to the line mentioned before).

Next we transform the area element. In cartesian coordinates it is $dxdy$ while in polar coordinates it is $rdrd\theta$ so we have $r^5\sin^2\thetadrd\theta=r^4\sin^2\theta dxdy.$

Next we use 4. to express the sine: $r^4\sin^2\theta dxdy=r^4\frac{y^2}{r^2}dxdy=r^2y^2dxdy$.

Finally, we use 5. $r^2y^2dxdy=y^2(x^2+y^2)dxdy$.

Therefore, the transformed integral is

$\int_0^{\pi/6}\int_0^{2\sec\theta}r^5\sin^2\theta dr d\theta=\int_0^2\int_0^{\frac{1}{\sqrt{4}}x}y^2(x^2+y^2).$