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

The transformed integral is

The region of integration is a triangle delimited by the axis, the vertical line , and the line .

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

1. and

2. .

The first condition says that the region of integrarion is within the "slice" of the xy plane between the polar angle of and in the first quadrant. To make sense of the second condition we will convert to the Cartesian coordinates. Remember that

3.

4.

5.

6.

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

Now using 3. to express we obtain

The left inequality just says that we start from the origin. The right inequality, after cancelling , yields . Since we are in the first quadrant has to be positive so this inequality can be rewritten as (taking the square root) . So the region is a triangle delimited by the slice of the first quadrant between and and by the vertical line (see figure).

In other words, this triangle is delimited by the x axis, by the vertical line x=2 and by the line This helps us to define the boundaries of integration: we have (from the origin up to x=2) and (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 while in polar coordinates it is so we have

Next we use 4. to express the sine: .

Finally, we use 5. .

Therefore, the transformed integral is

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints