$$int_-fracpi6^fracpi6left(int_0^6cos(3 heta) r dr ight) d heta$$
And I obtained an answer of $frac112pi$. At the finish of the difficulty, I got$$ frac14left(frac16pi + 6sinleft(fracpi6^2 ight) - frac14left(frac-pi6+6sin(-pi) ight) ight) $$
which need to be $frac112pi$ which I am unsure if it is best or not.
You are watching: Use a double integral to find the area of the region. one loop of the rose r = 6 cos(3θ)
$$ int_-fracpi6^fracpi6left(int_0^6cos(3 heta) r katifund.orgrm dr ight) katifund.orgrm d heta$$ $$=18int_-fracpi6^fracpi6cos^2(3 heta) katifund.orgrm d heta$$Because the integrand also is an also feature, we have$$36int_0^fracpi6cos^2(3 heta) katifund.orgrm d heta$$Now we deserve to usage the fact that$$int cos^2(ax) katifund.orgrm dx=fracx2+frac14asin(2ax)+C$$Wbelow in our case $a=3$, therefore$$36left(fracpi12+frac112sinleft(pi ight) ight)=frac36pi12=3pi$$
Thanks for contributing a response to katifund.orgematics Stack Exchange!Please be sure to answer the question. Provide details and also share your research!
But avoid …Asking for help, clarification, or responding to various other answers.Making statements based on opinion; ago them up through referrals or individual experience.
Use katifund.orgJax to format equations. katifund.orgJax reference.
To learn even more, check out our tips on creating great answers.
Post Your Answer Discard
By clicking “Article Your Answer”, you agree to our terms of service, privacy plan and cookie plan
Not the answer you're looking for? Browse various other questions tagged calculus integration or ask your own question.
website architecture / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. rev2021.9.10.40187
See more: Consulate Of New Zealand Consulate San Francisco, Consulate Of New Zealand San Francisco