Polar coordinates to evaluate double integral

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SUMMARY

The discussion focuses on evaluating the double integral $$\iint xy\, dxdy$$ over the region defined by the polar equation $$r=\sin(2\theta)$$ for $$0<\theta<\pi/2$$. Participants emphasize the importance of converting Cartesian coordinates to polar coordinates for this evaluation. A suggestion is made to sketch the region R enclosed by the curve to better understand the limits of integration. This approach is crucial for successfully setting up the integral in polar form.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Knowledge of polar coordinates and their conversion from Cartesian coordinates
  • Familiarity with the polar equation $$r=\sin(2\theta)$$
  • Ability to sketch curves defined by polar equations
NEXT STEPS
  • Learn how to convert Cartesian coordinates to polar coordinates for double integrals
  • Study the properties of polar curves, specifically $$r=\sin(2\theta)$$
  • Explore techniques for sketching regions defined by polar equations
  • Review the process of setting up double integrals in polar coordinates
USEFUL FOR

Students and educators in calculus, mathematicians focusing on integral calculus, and anyone looking to deepen their understanding of polar coordinates and double integrals.

brunette15
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I am trying to evaluate \int\int xy dxdy over the region R that is defined by r=sin(2theta), from 0<theta<pi/2. I am struggling on where to begin with this. I have tried converting to polar coordinates but am not really getting anywhere. Any guidance would be really appreciated (Crying)
 
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brunette15 said:
I am trying to evaluate $$\iint xy\, dxdy$$ over the region R that is defined by $$r=\sin(2\theta)$$, from $$0<\theta<\pi/2$$. I am struggling on where to begin with this. I have tried converting to polar coordinates but am not really getting anywhere. Any guidance would be really appreciated (Crying)
Converting to polar coordinates should certainly be the way to go. Can you show what you have done so far, so that we can see why it's "not really getting anywhere"?

The conditions $$r=\sin(2\theta)$$, from $$0<\theta<\pi/2$$, define a closed curve. Presumably the region R is meant to be the region enclosed by this curve. Drawing a rough sketch of R might help you to see what the limits of integration should be, for the polar coordinates.
 

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