Triple Integrals in Cartesian Coordinates

Calculation of a triple integral in Cartesian coordinates can be reduced to the consequent calculation of three integrals of one variable.

Consider the case when a three dimensional region U is a type I region, i.e. any straight line parallel to the z-axis intersects the boundary of the region U in no more than 2 points. Let the region U be bounded below by the surface z = z₁(x, y), and above by the surface z = z₂(x, y) (Figure 1).

A type I region bounded by two surfaces — Figure 1.

The projection of the solid \(U\) onto the \(xy\)-plane is the region \(D\) (Figure \(2\)).

We suppose that the functions \({z_1}\left( {x,y} \right)\) and \({z_2}\left( {x,y} \right)\) are continuous in the region \(D.\) Then for any function \(f\left( {x,y,z} \right)\) continuous in the region \(U\) we can write the relationship:

\[\iiint\limits_U {f\left( {x,y,z} \right)dV} = \iint\limits_D {\left[ {\int\limits_{{z_1}\left( {x,y} \right)}^{{z_2}\left( {x,y} \right)} {f\left( {x,y,z} \right)dz} } \right]dA}\]

Thus, calculating a triple integral is reduced to calculating a double integral, where the integrand is an one-dimensional integral. In the given case, we need first to calculate the inner integral with respect to the variable \(z,\) and then the double integral with respect to the variables \(x\) and \(y.\)

If \(D\left( {x,y} \right)\) is a type \(I\) region in the \(xy\)-plane (see Iterated Integrals), which is bounded by the lines

\[x = a,\;\;x = b,\;\; y = {f_1}\left( x \right),\;\; y = {f_2}\left( x \right),\]

where \({f_1}\left( x \right),\) \({f_2}\left( x \right)\) are continuous functions on the intervals \(\left[ {a,b} \right]\) and \({f_1}\left( x \right) \le {f_2}\left( x \right),\) then writing the double integral as an iterated integral, we get

\[\iiint\limits_U {f\left( {x,y,z} \right)dV} = \int\limits_a^b {dx} \int\limits_{{f_1}\left( x \right)}^{{f_2}\left( x \right)} {dy} \int\limits_{{z_1}\left( {x,y} \right)}^{{z_2}\left( {x,y} \right)} {f\left( {x,y,z} \right)dz}.\]

If \(D\left( {x,y} \right)\) is the region of type \(II\) and bounded by the lines

\[y = c,\;\;y = d,\;\; x = {\varphi _1}\left( y \right),\;\; x = {\varphi _2}\left( y \right),\]

where the functions \({\varphi _1}\left( y \right),\) \({\varphi _2}\left( y \right)\) are continuous on the interval \(\left[ {c,d} \right]\) so that \({\varphi _1}\left( y \right) \le {\varphi _2}\left( y \right),\) we can rewrite the formula for the triple integral as

\[\iiint\limits_U {f\left( {x,y,z} \right)dV} = \int\limits_c^d {dy} \int\limits_{{\varphi _1}\left( y \right)}^{{\varphi _2}\left( y \right)} {dx} \int\limits_{{z_1}\left( {x,y} \right)}^{{z_2}\left( {x,y} \right)} {f\left( {x,y,z} \right)dz} .\]

The last two formulas are an application of Fubini's theorem to triple integral. They enable us to reduce the computation of triple integrals to iterated integrals.

In the particular case when the region of integration \(U\) is the rectangular box \(\left[ {a,b} \right] \) \(\times \left[ {c,d} \right] \) \(\times \left[ {p,q} \right],\) the triple integral is given by

\[\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} = \int\limits_a^b {dx} \int\limits_c^d {dy} \int\limits_p^q {f\left( {x,y,z} \right)dz} .\]

Similar formulas for triple integrals exist for the solid regions \(U\) of type \(II\) or \(III.\) The projection of a solid region of type \(II\) is the region \(D\left( {y,z} \right)\) in the \(yz\)-plane, and, respectively, the projection of a solid region of type \(III\) is the region \(D\left( {x,z} \right)\) in the \(xz\)-plane.

Finally, if a solid region of integration \(U\) is more complex than considered above, we can break the region \(U\) into two or more smaller regions, each of which can be integrated separately.

Solved Problems

Example 1.

Evaluate the integral \[\int\limits_0^2 {\int\limits_0^z {\int\limits_0^y {xyzdxdydz} } } .\]

Solution.

Applying the Fubini's theorem, we can calculate the iterated integral starting from the inner one:

\[ I = \int\limits_0^2 {\int\limits_0^z {\int\limits_0^y {xyzdxdydz} } } = \int\limits_0^2 {dz} \int\limits_0^z {dy} \int\limits_0^y {xyzdz} = \int\limits_0^2 {dz} \int\limits_0^z {dy} \left[ {\left. {\left( {\frac{{{x^2}yz}}{2}} \right)} \right|_{x = 0}^{x = y}} \right] = \int\limits_0^2 {dz} \int\limits_0^z {\frac{{{y^3}z}}{2}dy} = \frac{1}{2}\int\limits_0^2 {dz} \int\limits_0^z {{y^3}zdy} = \frac{1}{2}\int\limits_0^2 {dz} \left[ {\left. {\left( {\frac{{{y^4}z}}{4}} \right)} \right|_{y = 0}^{y = z}} \right] = \frac{1}{2}\int\limits_0^2 {\frac{{{z^5}}}{4}dz} = \frac{1}{8}\int\limits_0^2 {{z^5}dz} = \frac{1}{8}\left. {\left( {\frac{{{z^6}}}{6}} \right)} \right|_0^2 = \frac{{64}}{{48}} = \frac{4}{3}.\]

Example 2.

Evaluate the integral \[\iiint\limits_U {\left( {1 - x} \right)dxdydz} ,\] where the region \(U\) lies in the first octant below the plane \[3x + 2y + z = 6.\]

Solution.

We rewrite the equation of the plane \(3x + 2y + z \) \(= 6\) in the form:

\[3x + 2y + z = 6,\;\; \Rightarrow \frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1,\]

The solid region of integration \(U\) is shown in Figure \(3.\)

Region U lying in the first octant below the plane 3x+2y+z=6 — Figure 3.

Projection of the region U on the xy-plane — Figure 4.

The limits of integrations for \(z\) range from \(z = 0\) to \(z = 6 - 3x - 2y.\) Considering the projection \(D\) in the \(xy\)-plane, we find that the variable \(y\) ranges from \(y = 0\) to \(y = 3 - {\frac{3}{2}} x\) (Figure \(4\)), while the variable \(x\) runs from \(0\) to \(2.\)

Consequently, the triple integral is expressed through iterated integral as

\[I = \iiint\limits_U {\left( {1 - x} \right)dxdydz} = \int\limits_0^2 {dx} \int\limits_0^{3 - \frac{3}{2}x} {dy} \int\limits_0^{6 - 3x - 2y} {\left( {1 - x} \right)dz} .\]

Calculate successively the three integrals to get

\[I = \int\limits_0^2 {dx} \int\limits_0^{3 - \frac{3}{2}x} {dy} \int\limits_0^{6 - 3x - 2y} {\left( {1 - x} \right)dz} = \int\limits_0^2 {dx} \int\limits_0^{3 - \frac{3}{2}x} {dy} \left[ {\left. {\left( {z - zx} \right)} \right|_{z = 0}^{z = 6 - 3x - 2y}} \right] = \int\limits_0^2 {dx} \int\limits_0^{3 - \frac{3}{2}x} {\left[ {6 - 3x - 2y - \left( {6 - 3x - 2y} \right)x} \right]dy} = \int\limits_0^2 {dx} \int\limits_0^{3 - \frac{3}{2}x} {\left( {6 - 3x - 2y - 6x + 3{x^2} + 2xy} \right)dy} = \int\limits_0^2 {dx} \int\limits_0^{3 - \frac{3}{2}x} {\left( {6 - 9x - 2y + 3{x^2} + 2xy} \right)dy} = \int\limits_0^2 {\left( {9 - 18x + \frac{{45}}{4}{x^2} - \frac{9}{4}{x^3}} \right)dx} = \left. {\left( {9x - \frac{{18}}{2}{x^2} + \frac{{45}}{{12}}{x^3} - \frac{9}{{16}}{x^4}} \right)} \right|_0^2 = 18 - 36 + 30 - 9 = 3.\]

See more problems on Page 2.