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For each of the following integrals, set up, but do not evaluate, the integral as an iterated integral as directed.

(a) Consider the iterated integral

sin(2Î¸ + 2y) dr dy.

Make a rough sketch of the region of integration for this iterated integral. Then set up the integral as an iterated integral with the opposite order of integration, i.e. in the order dy dz. Let D be the region inside the circle (x – 1)^2 + y^2 = 1, to the right of the line x = 3, and below the x-axis, shown below (shaded).

Set up the double integral âˆ«âˆ« (y + 3)^(4) as an iterated integral in polar coordinates.

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02:25

Evaluate the given integral by changing to polar coordinates: âˆ«âˆ«R (x + y) dA, where R is the region that lies to the left of the Y-axis between the circles x^2 + y^2 = 1 and x^2 + y^2 = 9.

Step:When transforming an integral âˆ«âˆ«R f(x, y) dA to polar coordinates, we must substitute for x and y in f(x, y), substitute for dA, and find polar limits for the double integral:

First of all, in polar coordinates, x = rcosÎ¸ and y = rsinÎ¸.

Therefore, the integral becomes: âˆ«âˆ«R (rcosÎ¸ + rsinÎ¸) r dr dÎ¸.

05:39

Compute the following integrals: âˆ«(1 + 2y)dA where D is the triangular region enclosed by the lines y = 0, x = 2, and x = 1; âˆ«âˆ«(4 – y)dA where R is the region in the first quadrant enclosed by the circle r^2 + y^2 = 4 (Use polar coordinates to calculate this integral).

06:36

Evaluate the given integral by changing to polar coordinates.$\iint_{R} \sin \left(x^{2}+y^{2}\right) d A,$ where $R$ is the region in the first quadrant between the circles with center the origin and radii 1 and 3

03:40

Evaluate the given integral by changing to polar coordinates.$$\begin{array}{l}{\iint_{R} \sin \left(x^{2}+y^{2}\right) d A, \text { where } R \text { is the region in the first quadrant }} \\ {\text { between the circles with center the origin and radii } 1 \text { and } 3}\end{array}$$

01:50

Set up one double integral to integrate the function z^2 + y^2 over the region in the plane between the cardioid r = 1 + cos(Î¸) and the circle of radius 2 centered at the origin, pictured below. Use polar coordinates; simplify the integrand as much as you can, and do not evaluate the integral.

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