##### just two multi qustioms 9 and 10

 Calculus Tutor: None Selected Time limit: 1 Day

Apr 28th, 2015

There is no 9 and 10, which do you need done?

Apr 28th, 2015

If it's 12 and 13,

The rose in #12.  The general form for any rose is r=asinntheta:

In this case, n=5. SO the only option is C.

13 to follow

Apr 28th, 2015

For 13. I like to look at the graph first, https://www.desmos.com/calculator/iesi0dusm4.

SO we need the area inside one petal, the formula we need is

SO in order to do that we need the theta values where our function equals zero.  0=sin2theta.  We take the first two values which are 0,pi

so then we find

A=int 1/2(sin^2(2theta))dtheta from 0 to pi

That becomes

```A=1/16 (4 theta-sin(4 theta)) from 0 to pi, which equals pi/4
```

Apr 28th, 2015

Now, I am reading that theta restriction as limiting our function to the first quadrant, but we could evaluate from 0 to pi/2 and get pi/8.  Can you tell me if this is a restriction or theta value based on previous problems?  It can be read both ways

Apr 28th, 2015

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Apr 28th, 2015
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Apr 28th, 2015
Dec 4th, 2016
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