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C4 Integration Area of Ellipse watch

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    How do I answer this question? I keep trying but the answer is wrong. BTW what's the range? Would appreciate if someone could help me w this.
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    (Original post by aynnihsan)
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    How do I answer this question? I keep trying but the answer is wrong. BTW what's the range? Would appreciate if someone could help me w this.
    What have you tried so far?
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    (Original post by SeanFM)
    What have you tried so far?
    Well, I substituted the y=2sinx with the -1 and 1 range... but that's the unshaded area. How do I get the area of the shaded one?
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    Well this seems like a question involving parametric equations, so the formula you normally use is:

    Area = The integral of y*dx/dt with respect to t. So maybe try that?
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    (Original post by aynnihsan)
    Well, I substituted the y=2sinx with the -1 and 1 range... but that's the unshaded area. How do I get the area of the shaded one?
    I'm not sure I understand what you are trying to say

    It looks symmetrical so you only have to worry about finding one of the shaded areas, then you can double it.

    If you integrate from the point furthest right to the point furthest left on the graph, then you will get the unshaded area + the shaded area, but that might still be useful.

    You know that at some point on the graph you reach the first co-ordinate where the shaded region is bounded, and then you reach the last point where the shaded region is bounded and you end up on the opposite side of the unshaded region, so you could use what you know about areas and how to find them to find the area of the shaded region.
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    could you use 1/2∫ r2dΘ ?
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    If you have found the area of the unshaded region, then just subtract it from the area of the whole ellipse, which is  \pi \text{ab} , where  \text{a} and  \text{b} are the semi minor and semi major axes.
 
 
 
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