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Great question! The inner loop begins at 2pi/3 when r first equals zero. The inner loop ends once r is next equal to zero which occurs at 4pi/3. The halfway point between these values is pi. Please also note your integral bounds in polar should typically go from a lesser to a greater number to maintain counterclockwise rotation. Hope this helps! Please subscribe to my channel for all the Calc videos you could ever need and spread the word!

⁠what do you mean by half way point? @@problemsolvedmathtutoring883

⁠is the value of the “halfway point” always halfway between the two bounds of integration of the inner loop? ​⁠​⁠@@problemsolvedmathtutoring883

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Thanks so much for your kind and thoughtful advice! I will be sure to incorporate it and hope the videos can be helpful to more people! Take good care!

Thanks so much! Please subscribe to my channel which has all of Calculus 2 made up of videos like this one. Spread the word and be well!

I actually never noticed that! Serendipity :) Thanks for watching! Please consider subscribing to my channel for videos on all of Calc 1 and 2!

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Thanks so much! If you watch to minute 5 of this video you will find these kinds of problems. Hope you enjoy watching and subscribe to my channel for tons more videos on Calc. Be well!

Right on! So happy it helped. Please consider subscribing and checking out all my Calc videos on my channel. Take good care!

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Thanks so much for watching and your kind words. I hope you will please consider subscribing to my channel for tons of videos like these. Take care!

😊 Thanks for watching! I hope the video was helpful. Please consider subscribing and sharing. Be well!

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Happy this helped! Please kindly consider subscribing to my channel...all of Calculus is there...and liking the video. Be well!

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My pleasure! Please subscribe to and check out my channel. There are tons of playlists of videos like this one. Many more to come! Be well and thanks for watching!

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Thanks so much for your kind comment! Depending on the function for r, one option would be to factor out the negative and treat the problem as normal. I am not sure how the bounds would be affected unless the exact problem is available. Thanks for the thoughtful question! All the best to you, and please subscribe, like, and share!

My pleasure!! Happy to help :) Kindly please subscribe, like, and share away! Be well.

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Thanks for your question. Can you please specify which problem you are referring to at a specific minute mark of the video? Thanks! Hope all is well.

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Thanks so much for the kind words @nirajvala3804! Please consider subscribing to my channel and liking the video when you get the chance. Take great care!

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​@@problemsolvedmathtutoring883already did

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Thanks for your comment! Yup! The answer in the video checks out, but I will triple check and get back to you if I get something different. Thanks again!

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Right on! Happy to help! Kindly please subscribe, like, and share so more people can rock their tests down the road :) Take good care!

Thanks so much for asking! The explanation is utilizing squeeze theorem with the special case of sina/a . What you are suggesting also works wonderfully as a use of squeeze theorem! Thanks again and take good care!

Thanks fpr your comment! The trig limits I use are the results of using the Squeeze Theorem. My apologies on any confusion. Be well and please subscribe, like, and share.