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Math Help - polar graphs

  1. #1
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    polar graphs

    i did this in high school and i remember understanding it.. but now i'm really rusty and a bit confused:

    in this particular problem set, i am given a graph, and below it, r= \sqrt{\theta} and i am asked to find the area of the shaded region

    i know the formula for area just fine, with the integral and all of that..but i'm confused as to how/why one knows that, for this particular graph, \ 0<\theta<2\pi (or equal to), like, how do you know that those are the endpoints when they are not given? our solutions manual just... says that they are there.

    any help is appreciated, thank you!
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    Quote Originally Posted by buttonbear View Post
    i did this in high school and i remember understanding it.. but now i'm really rusty and a bit confused:

    in this particular problem set, i am given a graph, and below it, r= \sqrt{\theta} and i am asked to find the area of the shaded region

    i know the formula for area just fine, with the integral and all of that..but i'm confused as to how/why one knows that, for this particular graph, \ 0<\theta<2\pi (or equal to), like, how do you know that those are the endpoints when they are not given? our solutions manual just... says that they are there.

    any help is appreciated, thank you!
    you have to look at the "shaded" region and remember the ray sweeps out one complete turn in 2\pi radians
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    but... it appears as though that interval is different for every graph, like, for instance: for  r= 4 + 3sin\theta , the solution says  -\pi/2 < \theta < \pi/2 and i really don't get why :/
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    like, for instance, how would i know what that interval was for the cardioid  r= 1 + cos\theta ? :/
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    Quote Originally Posted by buttonbear View Post
    but... it appears as though that interval is different for every graph, like, for instance: for  r= 4 + 3sin\theta , the solution says  -\pi/2 < \theta < \pi/2 and i really don't get why :/
    without knowing exactly what the problem says, I assume they are using symmetry to determine the area of the cardioid.

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(4+3\sin{\theta})^2}{2} \, d\theta gives half the area of the cardioid.

    the whole thing is ...

    2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(4+3\sin{\theta})^2}{2} \, d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (4+3\sin{\theta})^2 \, d\theta
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    Quote Originally Posted by buttonbear View Post
    like, for instance, how would i know what that interval was for the cardioid  r= 1 + cos\theta ? :/
    2\int_0^{\pi} \frac{(1+\cos{\theta})^2}{2} \, d\theta =  \int_0^{\pi} (1+\cos{\theta})^2 \, d\theta<br />

    you need to review your polar graphs.
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    so.. is that something that one has to memorize? like, where do they get the range of  -\pi/2  to  \pi/2? and, yes, it does look like they are doing that, because in the manual it says 2 x the integral.. and then the final answer involves just the integral from 0 to positive  \pi/2
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    honestly.. i would review if i could, but i don't have any of my high school notes, and i really find the textbook that my instructor uses to be very difficult to understand. i guess i'm looking for some basic guidelines to polar graphs, because i'm lost as of right now. so, is it fair to say that one uses pi when dealing with cosine, and pi/2 when dealing with sine? does this have anything to do with the period of the wave? i'm sorry that these are such naive questions, but, i have no one else to go to, at this point.
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  9. #9
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    Quote Originally Posted by buttonbear View Post
    so.. is that something that one has to memorize? like, where do they get the range of  -\pi/2  to  \pi/2? and, yes, it does look like they are doing that, because in the manual it says 2 x the integral.. and then the final answer involves just the integral from 0 to positive  \pi/2
    you just need to be familiar with the symmetry of some polar graphs and understand the use of symmetry to find area.

    note that you are not "required" to use symmetry to get the area ...

    \int_0^{2\pi} \frac{(4+3\sin{\theta})^2}{2} \, d\theta =  \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (4+3\sin{\theta})^2 \, d\theta   <br />
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    i'm trying my best to understand, but, even without using symmetry, i'm not sure how to get the endpoints for the integrals..

    like, another example, is i have a "4 petaled rose" graph of  r= sin2\theta
    and only the first petal is shaded.. but, i'm unsure of what i should use as the endpoints for that integral
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  11. #11
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    Quote Originally Posted by buttonbear View Post
    i'm trying my best to understand, but, even without using symmetry, i'm not sure how to get the endpoints for the integrals..

    like, another example, is i have a "4 petaled rose" graph of  r= sin2\theta
    and only the first petal is shaded.. but, i'm unsure of what i should use as the endpoints for that integral
    half a petal, doubled.

    \int_0^{\frac{\pi}{4}} \sin^2(2\theta) \, d\theta

    go here ...

    Calculus Videos - 41 - Polar Coordinates and Graphs
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    thank you so much for putting up with my absurd questions
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