I'm stuck with the question below. I'd highly appreciate for some steps. When I get the last integral respect to y, it gets really complicated.
${\int_0^4}{\int_{-2\sqrt{16-y^2}}^{2\sqrt{16-y^2}}} \int_0^{{(16-x^2)}/{(4-y^2)}} ~dzdxdy$
thank you
2026-03-29 03:20:21.1774754421
Triple integral question with long limits
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The inner integral evaluates to $\dfrac{16-x^2}{4-y^2}$. Then integrate this for $x$ from $-2\sqrt{16-y^2}$ to $2\sqrt{16-y^2}$, you get
$$\int_{-2\sqrt{16-y^2}}^{2\sqrt{16-y^2}}\dfrac{16-x^2}{4-y^2} \;\mathrm dx=\frac{[16x-x^3/3]_{-2\sqrt{16-y^2}}^{2\sqrt{16-y^2}}}{4-y^2}\\ =2\frac{[x(16-x^2/3)]_{0}^{2\sqrt{16-y^2}}}{4-y^2} =\frac{4\sqrt{16-y^2}}{4-y^2}\left(16-4/3(16-y^2)\right)\\ =\frac{16/3\cdot\sqrt{16-y^2}}{4-y^2}\left(12-16+y^2\right)=-16/3\cdot\sqrt{16-y^2}$$
And finally you integrate the last result for $y$ from $0$ to $4$. While computing the integral directly is not difficult, you may as well remark that, appart from the coefficient $-16/3$, it's the area of one fourth of a disc of radius $4$, that is $4\pi$, and your triple integral evaluates to:
$$-64\pi/3$$