Bruce wayne Asked a Question
May 7, 2020 4:12 pmpts 30 pts
2t r sin 0 de dr Ans. Ta Jo Jo 27t r sin 6 d0 dr Ans. TWa 3 Jo Jo rcos 8 dr de Jo Ans. T rdr de, over the area bounded between the circles r =2 cos 6 and r = 4 cos 6. Ans. 4 ul W.. vAD Cov We can a nS. t/2 pa (1 + cos 0) d de Ane 21T 4 sin 6 T Example
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  • Ruby negi
    this is the solution...
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  • Dhairya sharma
    solution with circle r=2cos(thitha)
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  • Abhishek singh Best Answer
    solution is attached, if you got confused in calculation, do it once by yourself, you will definately get the same answer. Best regards.
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    Bruce wayne
    thank u sir!
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