How to write $\iiint_R (x^2+y^2+z^2) dV $ where $R=\{x^2+y^2+z^2\leq 4, 1\leq z\leq 2\}$ the limits? I know that $1 \leq z\leq 2$ but for $x, y$ what should be?
Attempt: $y=\sqrt{4-z^2-x^2}$ or $x=\sqrt{4-z^2-y^2}$.
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1Do you know about spherical coordinates? – Gary Mar 01 '22 at 03:40
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@Gary how can I write it in that coordinates? Because the order is changed below – apa Mar 01 '22 at 03:49
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HINT
You can also apply the change of variables: \begin{align*} \begin{cases} x = r\cos(\theta)\\\\ y = r\sin(\theta)\\\\ z = z \end{cases} \end{align*}
Consequently, the region of integration is now given by: \begin{align*} R' = \left\{(r,\theta,z) : \left(0\leq r \leq \sqrt{4 - z^{2}}\right)\wedge(0\leq\theta\leq 2\pi)\wedge(1\leq z\leq 2)\right\} \end{align*}
Finally, one has that \begin{align*} I = \iiint_{R}(x^{2} + y^{2} + z^{2})\mathrm{d}V = \int_{1}^{2}\int_{0}^{2\pi}\int_{0}^{\sqrt{4-z^{2}}}r(r^{2} + z^{2})\mathrm{d}r\mathrm{d}\theta\mathrm{d}z \end{align*}
Can you take it from here?
Átila Correia
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$$\int\limits_{z=1}^2 dz \int\limits_{x = -\sqrt{4-z^2}}^{\sqrt{4-z^2}} dx \int\limits_{y = -\sqrt{4 - x^2-z^2}}^{\sqrt{4 - x^2-z^2}} dy\ f(x,y,z)$$
David G. Stork
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