z=9-x^3, y=-x^2+2, y=0, z=0, x is greater than and equal to 0.
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Originally Posted by asdf122345 z=9-x^3, y=-x^2+2, y=0, z=0, x is greater than and equal to 0. The definition of a triple integral is $\displaystyle \int_a^b dx \int_{g_1(x)}^{g_2(x)} dy \int_{h_1(x,y)}^{h_2(x,y)} dz $ By following this we get, $\displaystyle \int_0^{\sqrt{2}} dx \int_0^{-x^2 + 2} dy \int_0^{9-x^3} dz $
Last edited by AllanCuz; Apr 22nd 2010 at 05:31 PM.
Where the radical x come from?
Originally Posted by asdf122345 Where the radical x come from? $\displaystyle y=-x^2+2$ This gives our XY domain. To find our max value of x, set y=0. If you re-arrange such that $\displaystyle x^2 = 2 - y $ Clearly, x is max when y=0.
So shouldn't it be radical 2 instead of radical x? Or am I thinking something wrong.
Originally Posted by asdf122345 So shouldn't it be radical 2 instead of radical x? Or am I thinking something wrong. Yeah, i just wrote it wrong. It's radical 2
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