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P3 Jan 2002 Q8b watch

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    Hey guys - just doing some last final P3 revision for tmw, and I've come across this question. (btw: in my notation: a = parameter)

    x = cos a y = 4 sin a

    The question is: Find by integration the area enclosed by the ellipse.

    Basically, I've integrated -20 sin^2 a da to get 5 sin 2a - 10a. It's just the limits I need help on.
    Am I right in saying that my limits would be pi and 0? because the ellipse crosses the x-axis when y = 0. So I made 4 sin 0 = 0, and then when I inverse sin 0, i get 0 (and pi) for a.

    I put these limits (pi and 0) in and then got -20pi for the answer. In the markscheme it says that the answer is just 20pi. I'm assuming that you just make -20pi positive, since you can't have a negative area.

    Am I right? Any clarification would be much appreciated. Cheers.
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    (Original post by foowise)
    Hey guys - just doing some last final P3 revision for tmw, and I've come across this question. (btw: in my notation: a = parameter)

    x = cos a y = 4 sin a

    The question is: Find by integration the area enclosed by the ellipse.

    Basically, I've integrated -20 sin^2 a da to get 5 sin 2a - 10a. It's just the limits I need help on.
    Am I right in saying that my limits would be pi and 0? because the ellipse crosses the x-axis when y = 0. So I made 4 sin 0 = 0, and then when I inverse sin 0, i get 0 (and pi) for a.

    I put these limits (pi and 0) in and then got -20pi for the answer. In the markscheme it says that the answer is just 20pi. I'm assuming that you just make -20pi positive, since you can't have a negative area.

    Am I right? Any clarification would be much appreciated. Cheers.
    I'd say you were right, maybe your limits were upside down or the ellipse lies below the x axis.
 
 
 
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