## anonymous one year ago nick1234567 Area of one leaf of the rose r=3sin(4theta)

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1. Michele_Laino

I'm very sorry, it is 2:38 a.m. (Italy time zone) from me so I have to go to sleep

2. anonymous

3. anonymous

4. anonymous

Ur the best!

5. pooja195

ok any ideas on where to start?

6. anonymous

Nope no clue and it's my last prob due in 3 mins

7. anonymous

8. pooja195
9. anonymous

So? Area is?

10. anonymous

Only have 1min left thank u again

11. anonymous

Ideas???

12. anonymous

Needs suggestion in the ext 39 sec

13. pooja195

Things like this cant be done in that much time. You need to make sure you have time for it

14. pooja195

You will need to use $\huge~\rm~\int\limits_{}^{}1/2r^2d \emptyset$

15. IrishBoy123

if this is an emergency, this is a brief attempt: $$r=3sin(4 \theta)$$ $$r = 0, \ 4\theta = 0, \ \pi \implies \theta = 0, \pi/4, ....$$ $$A = \frac{1}{2} \int_{0}^{\pi / 4} 9 sin^2 (4 \theta) \ d \theta$$ $$= \frac{9}{4} \ \int_{0}^{\pi / 4} 1 - cos \ 8 \theta \ d \theta$$ $$= \frac{9}{4} \ [ \theta - \frac{1}{8}sin \ 8 \theta \ ]_{0}^{\pi / 4}$$ $$= \frac{9}{4} \ [ \frac{\pi}{4} ] = \frac{9 \pi}{16}$$

16. pooja195

^or that works too ;p

17. pooja195

You will need to use $\huge~\rm~\int\limits_{}^{}1/2r^2d \theta$ $\huge~\rm~\int\limits\limits_{\alpha }^{\beta }1/2r^2d \theta$ $\huge~\rm~\int\limits\limits\limits_{0 }^{\ \pi/4 }1/2(3\sin(4\theta))^2d \theta =\frac{ 9 }{ 16 }\pi$

18. Jack1

lol... why is this q orange?