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Thread: finding the area(problem with secant)

  1. #1
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    finding the area(problem with secant)

    this is the last one i have left on my homework. Find the area of the region inside r = 10 and to the right of r = 6*sec(theta)

    i'm having problem graphing the last one because the sec of 90 is zero right? which is totally throwing off my graph. what do i do in this situation.
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  2. #2
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    Area in polar coordinates

    Hello tomobson
    Quote Originally Posted by tomobson View Post
    this is the last one i have left on my homework. Find the area of the region inside r = 10 and to the right of r = 6*sec(theta)

    i'm having problem graphing the last one because the sec of 90 is zero right? which is totally throwing off my graph. what do i do in this situation.
    No, it's $\displaystyle \cos 90^o = 0$.

    But what you want is this:

    $\displaystyle r = 10$ is a circle, centre O, radius 10.

    $\displaystyle r = 6 \sec\theta$ is a straight line through through $\displaystyle (6, 0)$ perpendicular to the initial line. (Check it out: draw this line, and join any point on it to O, and then look at the value of $\displaystyle 6\sec\theta$.)

    So you need the area of the segment of the circle radius 10, cut off by a chord that is 6 units from the centre of the circle. (You get a 6, 8, 10 triangle, so it's very easy!)

    Grandad
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