Calculus · real student question

Evaluate the triple integral of sin of (x squared + y squared + z squared) to the power 3/2, over the region between the cone z = root of 3(x squared + y squared) and the sphere z = root of (1 - x squared - y squared).

Question

Evaluate Dxydxdy3(x2+y2)1x2y2sin ⁣((x2+y2+z2)3/2)dz.\iint_{D_{xy}}dx\,dy\int_{\sqrt{3(x^2+y^2)}}^{\sqrt{1-x^2-y^2}}\sin\!\left((x^2+y^2+z^2)^{3/2}\right)dz.

Step-by-step solution

  1. Identify the solid. The lower surface z=3rz=\sqrt{3}\,r (with r2=x2+y2r^2=x^2+y^2) is a cone, and the upper surface z=1r2z=\sqrt{1-r^2} is the upper unit hemisphere. They meet where 3r=1r2\sqrt3\,r=\sqrt{1-r^2}, i.e. 3r2=1r23r^2=1-r^2, so r=12r=\tfrac12: the region is the classic 'ice-cream cone' between the cone and the unit sphere, projecting onto the disc r12r\le\tfrac12.

  2. Choose spherical coordinates, not cylindrical. The integrand depends only on x2+y2+z2=ρ2x^2+y^2+z^2 = \rho^2, so in spherical coordinates it becomes sin(ρ3)\sin(\rho^3) - a function of one variable. Cylindrical coordinates would leave an inseparable sin ⁣((r2+z2)3/2)\sin\!\left((r^2+z^2)^{3/2}\right), which is why the conversion has to go all the way to spherical.

  3. Convert the cone into a constant polar angle. On the cone, tanφ=rz=13\tan\varphi = \dfrac{r}{z} = \dfrac{1}{\sqrt3}, so φ=π6\varphi = \dfrac{\pi}{6}. The solid is therefore 0ρ1,0φπ6,0θ2π,0\le\rho\le1,\qquad 0\le\varphi\le\frac{\pi}{6},\qquad 0\le\theta\le2\pi, with volume element dV=ρ2sinφdρdφdθdV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta.

  4. Separate the triple integral into three single ones. I=02π ⁣dθ0π/6 ⁣sinφdφ01ρ2sin(ρ3)dρ.I = \int_0^{2\pi}\!d\theta\int_0^{\pi/6}\!\sin\varphi\,d\varphi\int_0^{1}\rho^2\sin(\rho^3)\,d\rho. Each factor now involves a single variable.

  5. Evaluate each factor. 02πdθ=2π\int_0^{2\pi}d\theta = 2\pi; 0π/6sinφdφ=1cosπ6=132\int_0^{\pi/6}\sin\varphi\,d\varphi = 1-\cos\tfrac{\pi}{6} = 1-\tfrac{\sqrt3}{2}; and with u=ρ3u=\rho^3, du=3ρ2dρdu=3\rho^2 d\rho, 01ρ2sin(ρ3)dρ=1301sinudu=1cos13.\int_0^1\rho^2\sin(\rho^3)\,d\rho = \frac13\int_0^1\sin u\,du = \frac{1-\cos 1}{3}. The ρ2\rho^2 from the volume element is exactly what makes this substitution work.

  6. Multiply and check. I=2π(132)1cos132π(0.133975)(0.153244)0.128989.I = 2\pi\left(1-\frac{\sqrt3}{2}\right)\frac{1-\cos 1}{3} \approx 2\pi(0.133975)(0.153244) \approx 0.128989. Nested numerical quadrature of the original cylindrical form gives 0.12898921380.1289892138, matching to nine decimals.

Answer

2π3(132)(1cos1)0.128989\frac{2\pi}{3}\left(1-\frac{\sqrt3}{2}\right)\left(1-\cos 1\right) \approx 0.128989

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