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Math Help - Volume of ice cream cone

  1. #1
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    Volume of ice cream cone

    I need to find the volume of an ice cream cone bound by the hemisphere z = (8 = x^2 - y^2)^1/2 and the cone z = (x^2 + y^2)^1/2

    I know that is is a double integral but I can't even get started from there. Would someone please push or point me in the right direction? Thanks much for your effort. Frostking
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  2. #2
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    \sqrt{x^2+y^2}\leq z \leq \sqrt{8 - x^2 - y^2}

    The projection on the xy-plane, E, is where the two surfaces meet: \sqrt{x^2+y^2}=\sqrt{8 - x^2 - y^2} \Longleftrightarrow x^2+y^2=4

    So you can calculate it like this: \iint_E \left( \sqrt{8 - x^2 - y^2}-\sqrt{x^2+y^2}\right)dxdy
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  3. #3
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    To understand why we are doing this, consider this: \int_{\sqrt{x^2+y^2}}^{\sqrt{8 - x^2 - y^2}}\left(\iint_E dxdy\right)dz=\iint_E \left( \sqrt{8 - x^2 - y^2}-\sqrt{x^2+y^2}\right)dxdy
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  4. #4
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    So, if x^2 + y^2 is four then my bounds should be from 0 to 2 for both x and y is this correct? Frostking
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  5. #5
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    x^2+y^2=4 is a circle with the radius 2. So both x and y go from -2 to 2. I always switch to polar coordinates for problems like these though.

    Either go with -\sqrt{4-x^2}\leq y \leq\sqrt{4-x^2},\ -2\leq x\leq 2 or the other way around -\sqrt{4-y^2}\leq x \leq\sqrt{4-y^2},\ -2\leq y\leq 2
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