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Math Help - domain of two-variable, natural log function.

  1. #1
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    domain of two-variable, natural log function.

    I'm having trouble graphing the domain of the function f(x,y)=ln(9-x^2-9y^2). I don't need help drawing the graphg itself, I just need assurance that I'm on the right track.

    The domain should be all x,y such that 9-x^2-9y^2 \geq 0

    -9y^2 \geq x^2-9

    y^2 \leq 1-\frac{1}{9}x^2

    y\leq\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}

    So do I just need to sketch the region below the curves y=\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}

    ? It seems like there must be an easier way.
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    I'm having trouble graphing the domain of the function f(x,y)=ln(9-x^2-9y^2). I don't need help drawing the graphg itself, I just need assurance that I'm on the right track.

    The domain should be all x,y such that 9-x^2-9y^2 \geq 0

    -9y^2 \geq x^2-9

    y^2 \leq 1-\frac{1}{9}x^2

    y\leq\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}

    So do I just need to sketch the region below the curves y=\pm\sqrt{(1-\frac{1}{3}x)(1+\frac{1}{3}x)}

    ? It seems like there must be an easier way.
    Do you know what the ellipse \frac{x^2}{9} + y^2 \leq 1 looks like?
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    Do you know what the ellipse \frac{x^2}{9} + y^2 \leq 1 looks like?
    yeah, I see it now. I need to shade in the ellipse with center at origin, vertices at (0,\pm 1) and (\pm 3,0).
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