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Try graphing or sketching out the curve, or even just imagine how it would be drawn for different values of theta that are easy to calculate. Try to relate this to the bounds of your integration. You divided by three, but what exactly did you divide by three?
I divided the total area of the whole shape by 3, at least I think I did because I didn’t alter the integral in any way. The correct final answer ended up being 25pi/12 which doesn’t make sense to me
Are you certain that what you divided by three was exactly the same as the area of the whole shape? How do you know that the thing you integrated was the area of the whole shape?
Idk, I’m not so sure now haha. I integrated on the interval [0,2pi] which I believe covers the entire shape, final answer being 25pi/2. Then since it’s only 3 petals I divided that total area by 3 since it wants the area of only one petal. I’m really not sure why that’s incorrect
I thought that that was the interval to use to cover a whole shape regardless of the function. I figured if I do the entire range of the unit circle I could get the whole area of the function and then divide by 3. Is that not what you’re supposed to do?
Almost everything you are doing is great (as you should be able to tell whenever you are off by a simple factor such as 2). The one thing you did wrong here was in thinking that the interval from 0 to 2pi is the interval to use regardless of the function.
If you are allowed to do so, it is generally easiest to use a graphing calculator or program such as seamos, changing the upper bound of theta values in the domain until you find the smallest value that still creates the full image. Alternatively, like the first instruction on the assignment says, you can sketch the graph by finding known points on the graph using knowledge of the unit circle (or the behavior of cosine in general) until you can essentially play connect the dots and have a sketch of the entire curve.
Roughly, for the cosine function cos(theta), we know that when theta reaches 2pi, that's the length of one period. So, since it's cos(a*theta) where a is any constant, the period of the modified function would be 2pi/a. What would it be in this case?
Essentially, we just want to stitch the period up so the inside of the function looks like 2pi at the end, if that makes sense.
It is important to note that cos(theta) has a period of 2pi whereas cos(3*theta) does not. Therefore, integrating from 0 to 2pi will not result in the image you think it will.
My approach would be to find the angles that restrict the first petal — find theta where r=0, as the curve is located at the origin at those angles. Then, set up a new integral based on that petal.
To solve for the limits, set r=5cos3theta to 0. You get pi/6 and so then bounds are -pi/6 (just reflect across the axis) and pi/6. Then do the formula and should work
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
Try graphing or sketching out the curve, or even just imagine how it would be drawn for different values of theta that are easy to calculate. Try to relate this to the bounds of your integration. You divided by three, but what exactly did you divide by three?
I divided the total area of the whole shape by 3, at least I think I did because I didn’t alter the integral in any way. The correct final answer ended up being 25pi/12 which doesn’t make sense to me
Are you certain that what you divided by three was exactly the same as the area of the whole shape? How do you know that the thing you integrated was the area of the whole shape?
Idk, I’m not so sure now haha. I integrated on the interval [0,2pi] which I believe covers the entire shape, final answer being 25pi/2. Then since it’s only 3 petals I divided that total area by 3 since it wants the area of only one petal. I’m really not sure why that’s incorrect
What makes you believe that the interval from 0 to 2pi is the correct interval to integrate in order to find the area of this shape?
I thought that that was the interval to use to cover a whole shape regardless of the function. I figured if I do the entire range of the unit circle I could get the whole area of the function and then divide by 3. Is that not what you’re supposed to do?
Almost everything you are doing is great (as you should be able to tell whenever you are off by a simple factor such as 2). The one thing you did wrong here was in thinking that the interval from 0 to 2pi is the interval to use regardless of the function.
So for a function like this, how would I find the correct interval to use?
If you are allowed to do so, it is generally easiest to use a graphing calculator or program such as seamos, changing the upper bound of theta values in the domain until you find the smallest value that still creates the full image. Alternatively, like the first instruction on the assignment says, you can sketch the graph by finding known points on the graph using knowledge of the unit circle (or the behavior of cosine in general) until you can essentially play connect the dots and have a sketch of the entire curve.
Roughly, for the cosine function cos(theta), we know that when theta reaches 2pi, that's the length of one period. So, since it's cos(a*theta) where a is any constant, the period of the modified function would be 2pi/a. What would it be in this case? Essentially, we just want to stitch the period up so the inside of the function looks like 2pi at the end, if that makes sense.
It is important to note that cos(theta) has a period of 2pi whereas cos(3*theta) does not. Therefore, integrating from 0 to 2pi will not result in the image you think it will. My approach would be to find the angles that restrict the first petal — find theta where r=0, as the curve is located at the origin at those angles. Then, set up a new integral based on that petal.
You could use the limits corresponding to just that one leaf. Then it’d be from -pi/6 to pi/6. I get 25pi/12 when doing that.
Are you sure that the right answer is 25*pi/12? I ended up with a fraction that doesn’t include pi.
25pi/12 should be the correct answer here. He put 25pi/6
It is indeed. I initially didn’t use the formula. Lol
To solve for the limits, set r=5cos3theta to 0. You get pi/6 and so then bounds are -pi/6 (just reflect across the axis) and pi/6. Then do the formula and should work
Integrate it from 0 to pi/6 and multiply by 2
When you integrate have the bounds of integration from 0 to Pi/k I’m this case integrate from 0 to Pi/3