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Thread: area bounded by ∫∫ sin(theta)dA

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    area bounded by ∫∫ sin(theta)dA

    Evaluate area bounded by ∫∫ sin(theta)dA , where R is the region outside circle r =2 and inside cardroid r = 2 + 2 cos(theta) , in first octant ....
    I have sketched the area , is it correct ?

    The ans provided is 4/3 , however i gt 2 . Which part is wrong ? area bounded by ∫∫ sin(theta)dA-img_20160910_115045.jpgClick image for larger version. 

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ID:	36195area bounded by ∫∫ sin(theta)dA-img_20160910_115045.jpg
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    Re: area bounded by ∫∫ sin(theta)dA

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    Re: area bounded by ∫∫ sin(theta)dA

    it says the area outside the circle radius 2, and inside the cardoid.

    your picture shows the inside of the circle shaded.

    Further you say in the first octant but your picture shows the entire first quadrant included.

    Clean things up please.
    Thanks from xl5899
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by romsek View Post
    it says the area outside the circle radius 2, and inside the cardoid.

    your picture shows the inside of the circle shaded.

    Further you say in the first octant but your picture shows the entire first quadrant included.

    Clean things up please.
    after correction , my ans is 3 , not 8/3 , why am i wrong ?area bounded by ∫∫ sin(theta)dA-img_20160910_130753.jpg
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    Re: area bounded by ∫∫ sin(theta)dA

    do you know what an octant is ?
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by romsek View Post
    do you know what an octant is ?
    first octant = 0 to 90 degree , right ?
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by xl5899 View Post
    first octant = 0 to 90 degree , right ?
    that's quadrant, and octant is $\dfrac 1 8$ the unit circle.
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by romsek View Post
    do you know what an octant is ?
    u mean octant = pi / 8 ???
    If so , then the notes at b is wrong , right ?
    the shaded area is from 0 to pi / 2 , right ? the correct area should be (81/2) / 8 ?area bounded by ∫∫ sin(theta)dA-img_20160911_161001.jpg
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by xl5899 View Post
    u mean octant = pi / 8 ???
    An octant is $\displaystyle \frac{2\pi}{8}=\frac{\pi}{4}$ radians.
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by rubikwizard View Post
    An octant is $\displaystyle \frac{2\pi}{8}=\frac{\pi}{4}$ radians.
    an octant = a quadrant ?
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by xl5899 View Post
    an octant = a quadrant ?
    An octant is half of a quadrant. A quadrant is 90 degrees. An octant is 45 degrees.
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    Re: area bounded by ∫∫ sin(theta)dA

    Then, the notes that I uploaded in post #8 is wrong?
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    Re: area bounded by ∫∫ sin(theta)dA

    Remember that a full circle is $\displaystyle 2\pi$ radians (360 degrees).

    In your posts above I think you are using $\displaystyle \pi$ radians for a full circle.

    An octant is what you get when you divide a full circle into 8 parts.
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    Re: area bounded by ∫∫ sin(theta)dA

    Quote Originally Posted by rubikwizard View Post
    Remember that a full circle is $\displaystyle 2\pi$ radians (360 degrees).

    In your posts above I think you are using $\displaystyle \pi$ radians for a full circle.

    An octant is what you get when you divide a full circle into 8 parts.
    so , the notes is wrong ?
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    Re: area bounded by ∫∫ sin(theta)dA

    I haven't gone through the notes you gave in detail, but the replies above from romsek and myself are correct - an octant is half of a quadrant i.e. an octant = 45 degrees, or $\displaystyle \frac{2\pi}{8}$ radians. I still think that from what you have posted you are using a full circle to be $\displaystyle \pi$ radians instead of $\displaystyle 2\pi$ radians.
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