## TylerD one year ago @ganeshie8

1. TylerD

think my teacher might be wrong on something, need to check here. https://www.desmos.com/calculator/mdavtmnba2 for that graph, find the area inside the rose, but outside the circle.

2. TylerD

teachers solutions say just integrate (4cos(2theta))^2-2^2 from 0 to pi/6 then use symmetry.

3. TylerD

teacher used this equation here http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea_files/eq0008MP.gif but im not sure if u can do that in this case?

4. TylerD

|dw:1438082191447:dw|

5. ganeshie8

You want to find that area and multiply by 8 is it

6. TylerD

ya

7. TylerD

|dw:1438082672639:dw| but wouldnt that integral cut off there

8. ganeshie8

Oh you're right, my bad, lets try again

9. ganeshie8

$$\large \int\limits_{0}^{\pi/4}\frac{1}{2}(4\cos2\theta)^2~d\theta$$ gives below shaded area |dw:1438082959464:dw|

10. ganeshie8

$$\large \int\limits_{0}^{\pi/6}\frac{1}{2}(2)^2~d\theta$$ gives below shaded area |dw:1438083157606:dw|

11. TylerD

next one would be pi/6 to pi/4 for the cos(2theta)

12. ganeshie8

$$\large \int\limits_{\pi/6}^{\pi/4}\frac{1}{2}(4\cos2\theta)^2~d\theta$$ gives below shaded area |dw:1438083357364:dw|

13. ganeshie8

does that look okay now ? :)

14. TylerD

yep.

15. ganeshie8

lets work this and see if it matches with your teacher's answer

16. ganeshie8

just want you notice that we're adding and subtracting that last orange piece, which is simply a waste of work... evaluate the integrals and you will see what i mean

17. ganeshie8
18. ganeshie8

$$\large \int\limits_{0}^{\pi/\color{red}{6}}\frac{1}{2}(4\cos2\theta)^2~d\theta$$ gives below shaded area |dw:1438084278198:dw|

19. TylerD

dang, was hoping to get extra credit for proving it wrong.

20. ganeshie8

Haha! you had the right idea, pretty sure you will find other opportunities to catch ur teacher's mistakes ;)

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