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Me State Primary School third class teacher

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!

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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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​@@problemsolvedmathtutoring883already did

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@@ShanBojack Thanks so much for your kind words! Please share, like, and subscribe. Be well!

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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Thanks so much for your kind words and happy to help! Please check out my channel for more Calc videos and please feel free to share. Be well!

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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.

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Hi, he just simplified the bound from 9pi/6

he added both points of 11pie/6 and 7pie/6 then divided over 2 to find the midpoint

I am sorry, but I am not quite sure what you mean by this question. Thanks for asking it, though!

At 5:47, you found the value of theta by drawing a circle.

@@mady-son he set 2 + 4sin(theta) equal to 0 and solved for theta. I think he just drew the circle to help us visualize the unit circle!

Unit circle

did you figure it out? if yes, could you please help me? I have the same doubt

Thanks so much for your question! I was showing there is another option to use symmetry of the inner loop. By finding half the area of the inner loop which takes you from 7pi/6 to 3pi/2 and then doubling the result, you get the area of the entire inner loop. Hope this helps and thanks for watching!!!

@@problemsolvedmathtutoring883 so the top bound is 3pi/2 not pi/2 because r values are negative?

Correct, it has nothing to do with the negative r value but that 3pi/2 will sweep around half the inner loop. Great question! Be well!

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Happy to help Faten! Please subscribe to my channel and share it with others. I hope the videos continue to prove helpful to you! I have videos for every topic in Calculus 1 and 2!

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@@problemsolvedmathtutoring883Thanks and Already done sir🙂❤

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My pleasure, Stephanie! I hope you will check out my site, www.problemsolvedmath.com for full calculus courses! Be well and happy to help!

Digital hugs always welcome! Glad to help :) Be well!