Triple Integrals in Cartesian Coordinates
Solved Problems
Example 3.
Calculate the triple integral \[\iiint\limits_U {x{y^2}{z^3}dxdydz} ,\] where the region \(U\) (Figure \(5\)) is bounded by the surfaces
\[z = xy, y = x, x = 0, x = 1, z = 0.\]
Solution.
The projection of the solid region \(U\) onto the \(xy\)-plane looks as shown in Figure \(6.\)
Figure 5. Figure 6. Taking this into account, we find the corresponding iterated integral:
\[
I = \iiint\limits_U {x{y^2}{z^3}dxdydz}
= \int\limits_0^1 {dx} \int\limits_0^x {dy} \int\limits_0^{xy} {x{y^2}{z^3}dz}
= \int\limits_0^1 {dx} \int\limits_0^x {dy} \left[ {\left. {\left( {\frac{{x{y^2}{z^4}}}{4}} \right)} \right|_{z = 0}^{z = xy}} \right]
= \int\limits_0^1 {dx} \int\limits_0^x {\left( {x{y^2}\frac{{{x^4}{y^4}}}{4}} \right)dy}
= \frac{1}{4}\int\limits_0^1 {dx} \int\limits_0^x {{x^5}{y^6}dy}
= \frac{1}{4}\int\limits_0^1 {dx} \left[ {\left. {\left( {\frac{{{x^5}{y^7}}}{7}} \right)} \right|_{y = 0}^{y = x}} \right]
= \frac{1}{4}\int\limits_0^1 {\left( {{x^5}\frac{{{x^7}}}{7}} \right)dx}
= \frac{1}{{28}}\int\limits_0^1 {{x^{12}}dx}
= \frac{1}{{28}}\left. {\left( {\frac{{{x^{13}}}}{{13}}} \right)} \right|_0^1
= \frac{1}{{28}} \cdot \frac{1}{{13}} = \frac{1}{{364}}.\]
Example 4.
Express the triple integral \[\iiint\limits_U {dxdydz} \] in terms of iterated integrals in six different ways. The region \(U\) lies in the first octant and is bounded by the cylinder \({x^2} + {z^2} = 4\) and the plane \(y = 3\) (Figure \(7\)). Find the value of the integral.
Solution.
Figure 7. Figure 8. If the order of integration is \(z-y-x,\) then the iterated integral can be written as
\[{I_1} = \iiint\limits_U {dxdydz} = \int\limits_0^2 {dx} \int\limits_0^3 {dy} \int\limits_0^{\sqrt {4 - {x^2}} } {dz} .\]
Similarly, we can write the iterated integral to carry out the integration in the order \(z-x-y:\)
\[{I_2} = \int\limits_0^3 {dy} \int\limits_0^2 {dx} \int\limits_0^{\sqrt {4 - {x^2}} } {dz} .\]
Now we consider the case \(x-y-z,\) i.e. when the first inner integral is taken over the variable \(x.\) Then
\[{I_3} = \int\limits_0^2 {dz} \int\limits_0^3 {dy} \int\limits_0^{\sqrt {4 - {z^2}} } {dx} .\]
Since the projection of the solid onto the \(yz\)-plane is a rectangle (Figure \(8\)), then by changing the order of integration over \(y\) and \(z,\) we have
\[{I_4} = \int\limits_0^3 {dy\int\limits_0^2 {dz\int\limits_0^{\sqrt {4 - {z^2}} } {dx} } } .\]
Finally we can write the iterated integral in the order \(\left({y-x-z}\right)\) (starting from the inner integral):
\[{I_5} = \int\limits_0^2 {dz\int\limits_0^{\sqrt {4 - {z^2}} } {dx\int\limits_0^3 {dy} } } .\]
The last \(6\)th way looks as follows:
\[{I_6} = \int\limits_0^2 {dx\int\limits_0^{\sqrt {4 - {x^2}} } {dz\int\limits_0^3 {dy} } } .\]
We can use any of the iterated integrals to calculate the value of the initial triple integral. Taking the last one, we get
\[I = {I_6} = \int\limits_0^2 {dx\int\limits_0^{\sqrt {4 - {x^2}} } {dz\int\limits_0^3 {dy} } } = \int\limits_0^2 {dx\int\limits_0^{\sqrt {4 - {x^2}} } {dz \cdot \left[ {\left. y \right|_0^3} \right]} } = 3\int\limits_0^2 {dx\int\limits_0^{\sqrt {4 - {x^2}} } {dz} } = 3\int\limits_0^2 {dx\left[ {\left. z \right|_0^{\sqrt {4 - {x^2}} }} \right]} = 3\int\limits_0^2 {\sqrt {4 - {x^2}} dx} .\]
Make the substitution:
\[x = 2\sin t,\;\; \Rightarrow dx = 2\cos tdt,\]
\[x = 0,\;\; \Rightarrow t = 0,\]
\[x = 2,\;\; \Rightarrow \sin t = 1,\;\; \Rightarrow t = \frac{\pi }{2}.\]
As a result, we obtain:
\[I = 3\int\limits_0^2 {\sqrt {4 - {x^2}} dx} = 3\int\limits_0^{\frac{\pi }{2}} {\sqrt {4 - {{\left( {2\sin t} \right)}^2}} \cdot 2\cos tdt} = 12\int\limits_0^{\frac{\pi }{2}} {\sqrt {1 - {{\sin }^2}t} \cos tdt} = 12\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} = 12\int\limits_0^{\frac{\pi }{2}} {\frac{{1 + \cos 2t}}{2}dt} = 6\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}} = 6 \cdot \frac{\pi }{2} = 3\pi .\]
It is easy to check that this value is just \(\frac{1}{4}\) of the volume of the cylinder.
