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1. Given to the right is the graph of a portion of four curves: x = 0, y = 1, x – V3y = 0 and a? + y? = 4. Note that these curves divide the plane into 3 separate regions, which have been marked on the diagram.

1. Given to the right is the graph of a portion of four curves: x = 0, y = 1, x – V3y = 0 and a? + y? = 4. Note that these curves divide the plane into 3 separate regions, which have been marked on the diagram.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1. Given to the right is the graph of a portion of four curves: \(x = 0\), \(y = 1\), \(x - \sqrt{3}y = 0\) and \(x^2 + y^2 = 4\). Note that these curves divide the plane into 3 separate regions, which have been marked on the diagram. (a) Write \(\int_{R_1} 2x \, dA\) as an iterated integral, both in the “\(dx \, dy\)” order and in the “\(dy \, dx\)” order. Then, evaluate one of the two integrals. (b) Set up \(\int_{R_2} 2x \, dA\) as an iterated integral, both in the “\(dx \, dy\)” order and in the “\(dy \, dx\)” order. (c) Use polar coordinates to evaluate \(\int_{R_3} (x^2 + y^2) \, dA\). (d) Use polar coordinates to evaluate \(\int_{R} 12x \, dA\), where \(R\) is the region formed by combining the regions \(R_1\) and \(R_2\). (e) Set up, but do not evaluate, an iterated integral equivalent to \(\int_{S} 2x \, dA\), where \(S\) is the region formed by combining the regions \(R_2\) and \(R_3\). Use whatever coordinate system you think is easiest. **Diagram Explanation:** The diagram on the right shows a circle centered at the origin with a radius of 2. The circle is divided into three regions—\(R_1\), \(R_2\), and \(R_3\)—by the vertical line \(x = 0\), the horizontal line \(y = 1\), and the line \(x = \sqrt{3}y\). These regions lie within the bounds of the circle \(x^2 + y^2 = 4\).
Transcribed Image Text:1. Given to the right is the graph of a portion of four curves: \(x = 0\), \(y = 1\), \(x - \sqrt{3}y = 0\) and \(x^2 + y^2 = 4\). Note that these curves divide the plane into 3 separate regions, which have been marked on the diagram. (a) Write \(\int_{R_1} 2x \, dA\) as an iterated integral, both in the “\(dx \, dy\)” order and in the “\(dy \, dx\)” order. Then, evaluate one of the two integrals. (b) Set up \(\int_{R_2} 2x \, dA\) as an iterated integral, both in the “\(dx \, dy\)” order and in the “\(dy \, dx\)” order. (c) Use polar coordinates to evaluate \(\int_{R_3} (x^2 + y^2) \, dA\). (d) Use polar coordinates to evaluate \(\int_{R} 12x \, dA\), where \(R\) is the region formed by combining the regions \(R_1\) and \(R_2\). (e) Set up, but do not evaluate, an iterated integral equivalent to \(\int_{S} 2x \, dA\), where \(S\) is the region formed by combining the regions \(R_2\) and \(R_3\). Use whatever coordinate system you think is easiest. **Diagram Explanation:** The diagram on the right shows a circle centered at the origin with a radius of 2. The circle is divided into three regions—\(R_1\), \(R_2\), and \(R_3\)—by the vertical line \(x = 0\), the horizontal line \(y = 1\), and the line \(x = \sqrt{3}y\). These regions lie within the bounds of the circle \(x^2 + y^2 = 4\).
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