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Evaluate the triple integral where D is the region inside the cylinder $x^2 + y^2 = 1$ which is bounded below by the plane z=0 and bounded above by the plane 2x+4y+z=11.

I started this problem by changing into cylindrical coordinates and got this triple integral and was wondering if you can give me an okay before I start solving it. Thank you!

$$ \int_{-1}^1 \int_0^{2\pi} \int_0^{11-2rcos\theta -4rsin\theta} r(rcos\theta)^2 dzd\theta dr $$

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    $\begingroup$ $r$ should run from $0$ to $1$. Use 2-dimensional polar-coordinates. $\endgroup$ – Frieder Mar 19 '15 at 23:48
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The Jacobian of the cylindrical coordinates should be $r$.

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$$\begin{gathered} \int_{ - 1}^1 {\int_{ - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } {\int_0^{11 - 2x - 4y} {dzdydx = } } } \int_{ - 1}^1 {\int_{ - \sqrt {1 - {x^2}} }^{\sqrt {1 - {x^2}} } {(11 - 2x - 4y} )dydx} \hfill \\ = \int_0^1 {\int_0^{2\pi } {11 - 2\cos (\phi ) - 4sin(\phi )} rd\varphi dr} = 11\pi \hfill \\ \end{gathered}$$

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  • $\begingroup$ I don't understand how you went from a triple integral to a double integral? The answer is incorrect $\endgroup$ – Ayoshna Mar 19 '15 at 23:44
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    $\begingroup$ @Ayoshna: The function in triple integral is the constant function $1$, so $$\int\limits_o^{11 - 2x - 4y} {dz} = 11 - 2x - 4y$$ $\endgroup$ – Frieder Mar 20 '15 at 0:00

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