5.5 Triple integrals in cylindrical and spherical coordinates  (Page 5/12)

Now we can illustrate the following theorem for triple integrals in spherical coordinates with $({\rho }_{ijk}^{*},{\theta }_{ijk}^{*},{\phi }_{ijk}^{*})$ being any sample point in the spherical subbox $B_{ijk}.$ For the volume element of the subbox $\text{Δ}V$ in spherical coordinates, we have $\text{Δ}V=\left(\text{Δ}\rho \right)\left(\rho \text{Δ}\phi \right)\left(\rho \text{sin}\phi \text{Δ}\theta \right),,$ as shown in the following figure.

As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals. Fubini’s theorem takes the following form.

As stated before, spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Let us look at some examples before we consider triple integrals in spherical coordinates on general spherical regions.

The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that $dV$ and $dA$ mean the increments in volume and area, respectively. The variables $V$ and $A$ are used as the variables for integration to express the integrals.

The triple integral of a continuous function $f\left(\rho ,\theta ,\phi \right)$ over a general solid region

in ${ℝ}^3,$ where $D$ is the projection of $E$ onto the $\rho \theta$ -plane, is

In particular, if $D=\left\{\left(\rho ,\theta \right)|g_1\left(\theta \right)\le \rho \le g_2\left(\theta \right),\alpha \le \theta \le \beta \right\},$ then we have

Similar formulas occur for projections onto the other coordinate planes.

Setting up a triple integral in spherical coordinates

Set up an integral for the volume of the region bounded by the cone $z=\sqrt{3\left(x^2+y^2\right)}$ and the hemisphere $z=\sqrt{4-x^2-y^2}$ (see the figure below).

Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:

For the cone: $z=\sqrt{3\left(x^2+y^2\right)}$ or $\rho \text{cos}\phi =\sqrt{3}\rho \text{sin}\phi$ or $\text{tan}\phi =\frac{1}{\sqrt{3}}$ or $\phi =\frac{\pi }{6}.$

For the sphere: $z=\sqrt{4-x^2-y^2}$ or $z^2+x^2+y^2=4$ or ${\rho }^2=4$ or $\rho =2.$

Thus, the triple integral for the volume is $V\left(E\right)={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\varphi =0}{\overset{\phi =\pi \text{/}6}{\int }}{\displaystyle \underset{\rho =0}{\overset{\rho =2}{\int }}{\rho }^2\text{sin}\phi d\rho d\phi d\theta }}}.$

Got questions? Get instant answers now!

Got questions? Get instant answers now!