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Set up (evaluate later as an exercise) the integral ∬ R(2x−y)dA by changing to polar coordinates, where R is the region in the first quadrant enclosed by the circleˣ²+ʸ²=4 and the lines x=0 and y=x 3



User Vitor Reis
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Final answer:

To set up the integral ∫ R(2x-y)dA by changing to polar coordinates, we first need to identify the limits of integration and the corresponding equations for x and y in terms of polar coordinates. The region R is defined by the circle x²+y²=4 and the lines x=0 and y=x³. To find the limits of integration for r, we solve the equation x=r*cos(θ) for θ. Setting x to 0 gives us θ=π/2, and setting y to x³ gives us r*cos(θ)=(r*sin(θ))³, which simplifies to r=sin(θ). Therefore, the limits of integration for r are from 0 to sin(θ). To find the limits of integration for θ, we calculate the angle at which the line intersects the circle. This angle is given by θ=tan⁻¹(2/2)=π/4. Therefore, the limits of integration for θ are from 0 to π/4.

Step-by-step explanation:

To set up the integral ∫ R(2x-y)dA by changing to polar coordinates, we first need to identify the limits of integration and the corresponding equations for x and y in terms of polar coordinates. The region R is defined by the circle x²+y²=4 and the lines x=0 and y=x³. To find the limits of integration for r, we solve the equation x=r*cos(θ) for θ. Setting x to 0 gives us θ=π/2, and setting y to x³ gives us r*cos(θ)=(r*sin(θ))³, which simplifies to r=sin(θ). Therefore, the limits of integration for r are from 0 to sin(θ). To find the limits of integration for θ, we calculate the angle at which the line intersects the circle. This angle is given by θ=tan⁻¹(2/2)=π/4. Therefore, the limits of integration for θ are from 0 to π/4. Now that we have the limits of integration, we can express the differential area element dA in polar coordinates. The area element in rectangular coordinates is dA=dxdy, and we can express dx and dy in terms of dr and dθ using the polar coordinate conversions. We have dx=rcos(θ)dr-sin(θ)dθ and dy=rsin(θ)dr+cos(θ)dθ. Multiplying these expressions together and simplifying, we get dA=rdrdθ. Finally, we substitute the expressions for x and y in terms of polar coordinates and the differential area element into the double integral ∫∫R(2x-y)dA. The integrand becomes 2(rcos(θ)-rsin(θ)), and the differential area element becomes rdrdθ. The limits of integration for r are from 0 to sin(θ), and the limits of integration for θ are from 0 to π/4. Therefore, the integral becomes ∫∫(2rcos(θ)-2rsin(θ))rdrdθ over the region R in polar coordinates.

User Mateo Vial
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