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omeli [17]
3 years ago
14

Sketch the region of integration and convert the polar integral into cartesian integral or sum of cartesian integral. Don't eval

uate: integral^pi/6_0 integral^2sec theta_0 r^5sin 2 theta drd theta

Mathematics
1 answer:
sashaice [31]3 years ago
5 0

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).

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