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     \displaystyle f(x,y) = \sqrt{(4-x^2-y^2)(x^2+y^2-9)}

    I have found the domain to be  \displaystyle 4 \leq x^2+y^2 \leq 9

    So the region of the annulus, bounded by two concentric circles, centre the origin with radius 2 and 3.

    But for the level curves I have tried letting  \displaystyle \lambda = f(x,y), \lambda \in \mathbb{R}, but I can't seem to form a nice equation.

    The answer says the level curves are concentric circles, but I can't see how.
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    (Original post by NotNotBatman)
     \displaystyle f(x,y) = \sqrt{(4-x^2-y^2)(x^2+y^2-9)}

    I have found the domain to be  \displaystyle 4 \leq x^2+y^2 \leq 9

    So the region of the annulus, bounded by two concentric circles, centre the origin with radius 2 and 3.

    But for the level curves I have tried letting  \displaystyle \lambda = f(x,y), \lambda \in \mathbb{R}, but I can't seem to form a nice equation.

    The answer says the level curves are concentric circles, but I can't see how.
    Denote r^2=x^2+y^2 then under the root we have -(r^2-4)(r^2-9) = -[(r^2)^2-13r^2+36] hence 0=(r^2)^2-13r^2+36+\lambda^2

    So then we just have x^2+y^2=\dfrac{13\pm \sqrt{25-4\lambda^2}}{2} hence for |\lambda| \leq \frac{5}{2} we have circles center O with radius r
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    (Original post by RDKGames)
    Denote r^2=x^2+y^2 then under the root we have -(r^2-4)(r^2-9) = -[(r^2)^2-13r^2+36] hence 0=(r^2)^2-13r^2+36+\lambda^2

    So then we just have x^2+y^2=\dfrac{13\pm \sqrt{25-\lambda^2}}{2} hence for |\lambda| \leq 5 we have circles center O with radius r
    Thanks; have I made another mistake or should the discriminant be 25-4\lambda^2 ?
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    (Original post by NotNotBatman)
    Thanks; have I made another mistake or should the discriminant be 25-4\lambda^2 ?
    Yeah, forgot to type out the 4 it seems.
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    (Original post by NotNotBatman)
     \displaystyle f(x,y) = \sqrt{(4-x^2-y^2)(x^2+y^2-9)}

    I have found the domain to be  \displaystyle 4 \leq x^2+y^2 \leq 9

    So the region of the annulus, bounded by two concentric circles, centre the origin with radius 2 and 3.

    But for the level curves I have tried letting  \displaystyle \lambda = f(x,y), \lambda \in \mathbb{R}, but I can't seem to form a nice equation.

    The answer says the level curves are concentric circles, but I can't see how.
    Perhaps worth noting that f(x, y) = \sqrt{(4-r^2)(r^2-9)}, where r^2 = x^2 + y^2, and so f(x, y) depends only on r. It's immediate that f(x, y) = k must be a union of circles.
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    (Original post by DFranklin)
    Perhaps worth noting that f(x, y) = \sqrt{(4-r^2)(r^2-9)}, where r^2 = x^2 + y^2, and so f(x, y) depends only on r. It's immediate that f(x, y) = k must be a union of circles.
    Thank you.
 
 
 
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