[b][size=150]Triple integrals[/size][/b][br][br]Consider [math]w=f(x,y,z)[/math] defined on a closed and bounded domain [math]D[/math] in [math]\mathbb{R}^3[/math]. Then [math]D[/math] can be contained in a cuboid [math][a_1,b_1]\times [a_2,b_2]\times [a_3,b_3]=\left\{(x,y,z) \ | \ a_1\leq x \leq b_1, \ a_2\leq y \leq b_2, \ a_3\leq z \leq b_3 \right\}[/math]. We divide the cuboid into small sub-cuboids by planes parallel to xy, yz, and xz planes. Choose the small sub-cuboids that are contained in [math]D[/math] and label them from [math]k=1[/math] to [math]k=N[/math]. For the [math]k^{\text{th}}[/math] sub-cuboid, we denote its side lengths to be [math]\Delta x_k, \Delta y_k[/math], and [math]\Delta z_k[/math]. Then its volume [math]\Delta V_k[/math] is [math]\Delta x_k \Delta y_k \Delta z_k[/math]. Moreover, we select an arbitrary point [math](x_k^*,y_k^*,z_k^*)[/math] in the sub-cuboid. [br][br]The [b]triple integral[/b] of [math]f[/math] over [math]D[/math] is defined to be the limit of the following "Riemann sum":[br][br][math]\iiint_D f(x,y,z) \ dV=\lim_{N\to\infty} \sum_{k=1}^N f(x_k^*,y_k^*,z_k^*)\Delta V_k[/math][br][br]The diagram shown below is the limiting procedure of cutting the domain [math]D[/math] into smaller and smaller sub-cuboids.[br]

[u]Physical interpretation of triple integral[/u][br][br]Suppose [math]D[/math] is a 3D solid and its density varies at different points of the solid according to [math]f(x,y,z)[/math]. Then [math]\iiint_D f(x,y,z) \ dV=[/math] Total mass of [math]D[/math].[br][br]In particular, if [math]f(x,y,z)=1[/math], i.e. the solid has "uniform density" of 1, we can regard the triple integral [math]\iiint_D 1 \ dV[/math] as the volume of the solid.[br][br][br][br]Suppose [math]D=[a,b]\times[c,d]\times[k,l][/math], i.e. a cuboid. We can calculate the triple integral using the following version of Fubini's theorem:[br][br][u]Fubini's theorem[/u]: Suppose [math]w=f(x,y,z)[/math] and [math]D=[a,b]\times[c,d]\times[k,l][/math]. Then we have[br][br][math]\iiint_D f(x,y,z) \ dV = \int_a^b\left(\int_c^d\left(\int_k^l f(x,y,z) \ dz\right) \ dy \right) \ dx[/math][br][br]This integral is also equal to any of the other five possible orderings for the iterated triple integral.[br][br][br][u]Example[/u]: Find the value of [math]\iiint_G xy^2z^3 \ dV[/math], where [math]G=[0,2]\times[-1,0]\times[0,2][/math].[br][br][u]Answer[/u]:[br][br][math]\iiint_G xy^2z^3 \ dV=\int_0^2\left(\int_{-1}^0\left(\int_0^2 xy^2z^3 \ dz\right) \ dy \right) \ dx[/math][br][math]=\int_0^2\left(\int_{-1}^0\left[\frac{xy^2z^4}4\right]_0^2 \ dy \right) \ dx=\int_0^2\left(\int_{-1}^0 4xy^2 \ dy \right) \ dx[/math][br][math]=\int_0^2 \left[\frac{4xy^3}3\right]_{-1}^0 \ dx = \int_0^2 \frac{4x}3 \ dx =\left[ \frac{4x^2}6\right]_0^2=\frac 83[/math][br][br][br][br]

[b][size=150]Triple integrals over a general domain[br][/size][/b][br]Suppose [math]w=f(x,y,z)[/math] is defined on a domain [math]D[/math], which is solid bounded by the graph of [math]z=G(x,y)[/math] from below and by the graph of [math]z=H(x,y)[/math] from above over a region [math]R[/math] on the xy plane. In other words, [math]H(x,y)\geq G(x,y)[/math] for all [math](x,y)[/math] in [math]R[/math]. This kind of domains is called a [b]simple xy-solid[/b]. We can reduce the triple integral over a simple xy-solid to the double integral as follows:[br][br][math]\iiint_D f(x,y,z) \ dV=\iint_R\left(\int_{G(x,y)}^{H(x,y)} f(x,y,z) \ dz\right) \ dA[/math][br][br]Moreover, if [math]R[/math] is a Type I region bounded from below by the graph of [math]y=g(x)[/math] and from above by the graph of [math]y=h(x)[/math] over [math][a,b][/math], we can express the triple integral as the following iterated integral:[br][br][math]\iiint_D f(x,y,z) \ dV=\int_a^b \left( \int_{g(x)}^{h(x)} \left(\int_{G(x,y)}^{H(x,y)} f(x,y,z) \ dz\right) \ dy \right) \ dx[/math][br][br]We have similar results for [b]simple xz-solid[/b] and [b]simple yz-solid[/b]:[br][br][math]E[/math] is a simple xz-solid if it is bounded by the surfaces [math]y=P(x,z)[/math] and [math]y=Q(x,z)[/math] over the region [math]S[/math] on xz plane and [math]P(x,z)\leq Q(x,z)[/math] for all [math](x,z)[/math] in [math]S[/math]. Then the triple integral over [math]E[/math] is as follows:[br][br][math]\iiint_E f(x,y,z) \ dV=\iint_S\left(\int_{P(x,z)}^{Q(x,z)} f(x,y,z) \ dy \right) \ dA[/math][br][br](The definition of simple yz-solid and its analogous result are left as exercise.)[br][br][br]

[u]Example[/u]: Let [math]G[/math] be the wedge in the first octant i.e. the set of points in [math]\mathbb{R}^3[/math] such that x,y,z coordinates are non-negative, that is cut from the cylindrical solid [math]y^2+z^2\leq 1[/math] by the planes [math]y=x[/math] and [math]x=0[/math]. Find the value of the triple integral [math]\iiint_G z \ dV[/math].[br][br][u]Answer[/u]:[br][br]The domain [math]G[/math] can be regarded as a simple xy-solid such that it is bounded by the cylinder [math]y^2+z^2=1[/math] from above and the plane [math]z=0[/math] from below over a triangular region [math]R[/math] as shown in the applet below. Notice that [br][br][math]y^2+z^2=1 \Rightarrow z=\sqrt{1-y^2}[/math] (Note: Positive square root is used because [math]z[/math] is non-negative)[br][br]Then, we have[br][br][math]\iiint_G z \ dV = \iint_R\left(\int_0^{\sqrt{1-y^2}} z \ dz \right) \ dA[/math][br][br]Moreover, [math]R[/math] can be regarded as a Type II region bounded by [math]x=0[/math] and [math]x=y[/math]. Therefore, we can express the triple integral as the following iterated integral:[br][br][math]\iiint_G z \ dV = \int_0^1\left(\int_0^y \left(\int_0^{\sqrt{1-y^2}} z \ dz \right) \ dx \right) \ dy[/math][br][math]=\int_0^1\left(\int_0^y \left[\frac{z^2}2\right]_0^{\sqrt{1-y^2}} \ dx \right) \ dy=\int_0^1\left(\int_0^y \frac{1-y^2}2 \ dx \right) \ dy[/math] [br][math]=\int_0^1\left[\frac{1-y^2}2 x\right]_0^y \ dy =\int_0^1 \frac{y-y^3}2 \ dy[/math][br][math]=\left[\frac{y^2}4-\frac{y^4}8\right]_0^1=\frac 18[/math]

[u]Example[/u]: Find the volume of the solid [math]G[/math] bounded by the two paraboloids [math]y=x^2+z^2[/math] and [math]y=16-3x^2-z^2[/math].[br][br][u]Answer[/u]:[br][br]Volume of [math]G[/math] can be computed by evaluating the triple integral [math]\iiint_G 1 \ dV[/math]. And [math]G[/math] can be regarded as a simple xz-solid bounded by [math]y=x^2+z^2[/math] and [math]y=16-3x^2-z^2[/math]. In order to find the region [math]R[/math] on which the solid is bounded by these two surfaces, we need to find the projection of the intersection of these two surfaces to the xz plane.[br][br]First, we combine the two equations of the surfaces. We have[br][br][math]x^2+z^2=16-3x^2-z^2 \Rightarrow 2x^2+z^2=8[/math][br][br]i.e., the projection of the intersection onto xz plane is a ellipse, as shown in the applet below. Therefore, [math]R[/math] is the region bounded by the ellipse, which can be regarded as a Type I region bounded by [math]z=\sqrt{8-2x^2}[/math] and [math]z=-\sqrt{8-2x^2}[/math] for [math]-2\leq x \leq 2[/math]. Hence, we can write the triple integral as the following iterated integral:[br][br][math]\iiint_G 1 \ dV=\iint_R\left(\int_{x^2+z^2}^{16-3x^2-z^2} 1 \ dy\right) \ dA[/math][br][math]=\int_{-2}^2\left(\int_{-\sqrt{8-2x^2}}^{\sqrt{8-2x^2}}\left(\int_{x^2+z^2}^{16-3x^2-z^2} 1 \ dy \right) \ dz \right) \ dx[/math][br][math]=\int_{-2}^2\left(\int_{-\sqrt{8-2x^2}}^{\sqrt{8-2x^2}}(16-4x^2-2z^2) \ dz\right) \ dx[/math][br][math]=\int_{-2}^2\left[16z-4x^2z-\frac 23 z^3 \right]_{-\sqrt{8-2x^2}}^{\sqrt{8-2x^2}} \ dx[/math][br][math]=\frac{16\sqrt{2}}3\int_{-2}^2 (4-x^2)^{\frac 32} \ dx=\frac{32\sqrt{2}}3\int_{-\frac{\pi}2}^{\frac{\pi}2}\cos^4(\theta) \ d\theta[/math] (Note: Use substitution [math]x=2\sin(\theta)[/math])[br][math]=32\sqrt{2}\pi[/math][br]

[u]Changing the order of integration[/u][br][br]Sometimes it is necessary to change the order of integration in an iterated integral so that the it can be evaluated more easily. [br][br][u]Example[/u]: Suppose [math]E[/math] is a 3D solid such that [br][br][math]\iiint_E 12y^2z^3\sin(x^4) \ dV=\int_0^{\pi^{\frac 14}}\left(\int_0^z\left(\int_y^z 12y^2z^3\sin(x^4) \ dx\right) \ dy \right) \ dz[/math][br][br]Describe the solid [math]E[/math] and then change the order of integration in the iterated integral to evaluate the triple integral.[br][br][br][u]Answer[/u]:[br][br]Since the innermost integral is with respect to [math]x[/math], the solid [math]E[/math] is a simple yz-solid bounded by [math]x=y[/math] and [math]x=z[/math]. Moreover, the integrals with respect to [math]y[/math] and [math]z[/math] suggest that the simple yz-solid is over the region [math]R[/math] on yz plane, where [math]R[/math] is the region bounded by [math]y=0[/math] and [math]y=z[/math] for [math]0\leq z\leq \pi^{\frac 14}[/math]. The applet below shows that [math]E[/math] is a tetrahedron bounded by four planes: [math]x=y[/math], [math]x=z[/math], [math]y=0[/math], and [math]z=\pi^{\frac 14}[/math].[br][br][br]We can view [math]E[/math] as a simple xz-solid bounded by [math]y=x[/math] and [math]y=0[/math] over the triangular region bounded by [math]x=0[/math] and [math]x=z[/math] for [math]0\leq z \leq \pi^{\frac 14}[/math]. Therefore, we can write the iterated integral in a different order as follows:[br][br][math]\iiint_E 12y^2z^3\sin(x^4) \ dV=\int_0^{\pi^{\frac 14}}\left(\int_0^z\left(\int_0^x 12y^2z^3\sin(x^4) \ dy \right) \ dx \right) \ dz[/math][br][math]=\int_0^{\pi^{\frac 14}}\left(\int_0^z 4x^3z^3\sin(x^4) \ dx \right) \ dz[/math][br][math]=\int_0^{\pi^{\frac 14}}\left[-z^3\cos(x^4)\right]_0^z \ dz[/math][br][math]=\int_0^{\pi^{\frac 14}} (-z^3\cos(z^4)+z^3) \ dz[/math][br][math]=\left[\frac 14\sin(z^4)+\frac 14z^4\right]_0^{\pi^{\frac 14}}=\frac{\pi}4[/math][br][br][br]