Christmas number 6

$\begin{aligned} I & = \int_{-2}^2 \int_{-\sqrt{\frac {4-x^2}2}}^{\sqrt{\frac {4-x^2}2}} \int_{x^2+3y^2}^{8-x^2-y^2} dz \ dy \ dx \\ & = \int_{-2}^2 \int_{-\sqrt{\frac {4-x^2}2}}^{\sqrt{\frac {4-x^2}2}} z \ \bigg|_{x^2+3y^2}^{8-x^2-y^2} dy \ dx \\ & = \int_{-2}^2 \int_{-\sqrt{\frac {4-x^2}2}}^{\sqrt{\frac {4-x^2}2}} 8-2x^2-4y^2 \ dy \ dx \\ & = \int_{-2}^2 (8-2x^2)y- \frac {4y^3}3 \ \bigg|_{-\sqrt{\frac {4-x^2}2}}^{\sqrt{\frac {4-x^2}2}} \ dx \\ & = \int_{-2}^2 (8-2x^2)\cdot 2 \left(\frac {4-x^2}2 \right)^\frac 12 - \frac 43\cdot 2 \left(\frac {4-x^2}2 \right)^\frac 32 \ dx \\ & = \int_{-2}^2 8 \left(\frac {4-x^2}2 \right)^\frac 32 - \frac 83 \left(\frac {4-x^2}2 \right)^\frac 32 \ dx \\ & = \int_{-2}^2 \frac {16}3 \left(\frac {4-x^2}2 \right)^\frac 32 \ dx \\ & = \int_{-2}^2 \frac {64}{3\sqrt 2} \left({\color{#3D99F6}1-\frac {x^2}4} \right)^\frac 32 \ dx & \small \color{#3D99F6} \text {Let } \sin \theta = \frac x2, \ dx = 2\cos \theta \ d \theta \\ & = \frac {128}{3\sqrt 2} \int_{-\frac \pi 2}^\frac \pi 2 {\color{#3D99F6}\cos^4 \theta} \ d \theta & \small \color{#3D99F6} \text{The integral is even.} \\ & = \frac {\color{#3D99F6}256}{3\sqrt 2} \int_{\color{#3D99F6}0}^\frac \pi 2 \sin^0 \theta \ \cos^4 \theta \ d \theta \\ & = \frac {128}{3\sqrt 2} B \left(\frac 12, \frac 52 \right) & \small \color{#3D99F6} B(m,n) \text{ is the beta function.} \\ & = \frac {128}{3\sqrt 2} \cdot \frac {\Gamma \left(\frac 12 \right)\Gamma \left(\frac 52 \right)}{\Gamma \left(3 \right)} & \small \color{#3D99F6} \Gamma (n) \text{ is the gamma function.} \\ & = \frac {128}{3\sqrt 2} \cdot \frac {\sqrt \pi \cdot \frac 32 \cdot \frac 12 \Gamma \left(\frac 12 \right)}{2!} \\ & = \frac {16 \pi}{\sqrt 2} \approx \boxed{35.543} \end{aligned}$

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Reference:

• Beta function
• Gamma function

The integral is equal to $\begin{aligned} I & = \iint_{x^2 + 2y^2 \le 4} (8 - 2x^2 - 4y^2)\,dx\,dy \; = \; 2\iint_{x^2 + 2y^2 \le 4}(4 - x^2 - 2y^2)\,dx\,dy \\ & = \sqrt{2}\iint_{X^2+Y^2 \le 4} (4 - X^2 - Y^2)\,dX\,dY \; = \; \sqrt{2}\int_0^2 r\,dr \int_0^{2\pi} (4 - r^2)\,d\theta \\ & = 2\pi\sqrt{2}\int_0^2 (4r - r^3)\,dr \; = \; 2\pi\sqrt{2}\Big[2r^2 - \tfrac14r^4\Big]_0^2 \; = \; 8\pi\sqrt{2} \end{aligned}$ making the answer $\boxed{35.543063505}$ . Here we have first performed the integration with respect to $z$ , recognised what is left as the integral over a region in $\mathbb{R}^2$ , introduced a change of variables $X = x$ , $Y = y\sqrt{2}$ to convert the region of integration to a circle, and then introduced polar coordinates $r,\theta$ to complete the integral.