My math problem is to evaluate the integral
$$\iiint_{D}cos(x+2y+3z) dV $$
where $D=\left \{ (x,y,z):x^2+y^2+z^2\leq 1 \right \}$
I tried approaching it with spherical coordinate by substituting $x=\rho \space \sin\phi \space \cos\theta$, $y=\rho \space \sin\phi \space \sin\theta$, $z=\rho\space \cos\phi $, but it doesn't help much since the function inside the integral ends up being more complicated. (cosine and sine functions inside cosine)
I also tried using Cartesian coordinate, and changing the integral into
$$8\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\int_{0}^{\sqrt{1-x^2-y^2}} \cos(x+2y+3z) \space \space dzdydx $$
$$=8\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}} \frac{\sin(x+2y+3\sqrt{1-x^2-y^2})-\sin(x+2y)}3 \space \space dydx$$
But I have no idea how to evaluate it, and I don't have any more ideas on this problem.
I see someone else in another post saying we can make use of spherical symmetry and "replace the integrand by $\cos(hz)$, where $h=|(1,2,3)|$. Then, integrate in slices perpendicular to the $z$-axis.", but I have no idea if it would work or why it would work. (and I don't have enough reputation points to comment on that post 🙁 )
Any help is appreciated. Thanks!
Best Answer
Align the $x$-axis along $(1,2,3)$, the normal direction of the plane $x+2y+3z$, by rotating the angle $\alpha$ around the $z$-axis, followed by the angle $\beta$ around the $y$-axis, with $ \sin\alpha = \frac2{\sqrt5}$ and $ \sin\beta = \frac {3}{\sqrt{14}}$. In the new coordinates, we have
$$u = \frac1{\sqrt{14}}(x+2y+3z), \>\>\>\>\> D=\left \{ u^2+v^2+w^2 < 1 \right \}$$
The integral then simplifies and can be readily integrated in spherical coordinates as follows \begin{align} &\int_{D}\cos(x+2y+3z) \ dV \\ =&\int_{D}\cos(\sqrt{14}u)\ dV =\ 2\pi\int_0^1 \int_0^\pi \cos\left(\sqrt{14}r\cos\theta\right)r^2\sin\theta \ dr d\theta \\ = &\ \frac{4\pi}{\sqrt{14}} \int_0^1 \sin(\sqrt{14}r)\ rdr = \frac{2\pi}{7}\left(\frac{\sin \sqrt{14}}{\sqrt{14}}- \cos\sqrt{14}\right) \end{align}