Evaluation of Triple Integrals

Section 6.2 Evaluation of Triple Integrals

As with double integrals, triple integrals can be evaluated by computing iterated integrals if the domain is given in a suitable form. The form is

\begin{equation} D=\bigl\{(x,y,z)\colon a\leq x\leq b, \varphi(x)\leq y\leq\Phi(x), \psi(x,y)\leq z\leq\Psi(x,y)\bigr\}\text{.}\tag{6.2} \end{equation}

Note that the projection of \(D\) along the \(z\)-axis onto the \(xy\)-axis is given by

\begin{equation*} D_0:=\bigl\{(x,y,z)\colon a\leq x\leq b, \varphi(x)\leq y\leq\Phi(x)\}, \end{equation*}

and that the condition \(\psi(x,y)\leq z\leq\Psi(x,y)\) indicates that \(D\) is between the graph of \(\psi\) and \(\Psi\) over \(D_0\text{.}\) By slicing the domain into thin domains similarly as done in Section 5.2 one can show the following generalised version of Fubini’s theorem.

Theorem 6.2. Fubini’s Theorem.

Suppose that \(D\) has the form (6.2) with \(\varphi,\Phi\) continuous on \([a,b]\text{,}\) and \(\psi,\Psi\) continuous on \(D_0\) as defined above. If \(f\) is continuous on \(D\) then

\begin{equation*} \iiint_{D}f(x,y,z)\,dx\,dy\,dz =\int_a^b\Bigl( \int_{\varphi(x)}^{\Phi(x)} \Bigl(\int_{\psi(x,y)}^{\Psi(x,y)} f(x,y,z)\,dz \Bigr)\,dy\Bigr)\,dx\text{.} \end{equation*}

Note that we can interchange the roles of \(x,y,z\) arbitrarily.

Example 6.3.

Let \(D\) be the domain bounded below by \(z=\sqrt{x^2+y^2}\text{,}\) and bounded from above by \(9x^2+4y^2+z^2=5\text{.}\) Find the limits for the triple integral of a function \(f(x,y,z)\) over the domain \(D\text{.}\)

Solution.

We first find the projection, \(D_0\text{,}\) of the domain onto the \(xy\)-plane. The intersection of \(z=\sqrt{x^2+y^2}\) and \(9x^2+4y^2+z^2=5\) forms a curve in space. Its projection onto the \(xy\)-plane is the boundary of \(D_0\text{.}\) To find that projection we eliminate \(z\) from the two equations. If we substitute the first into the second equation we get

\begin{equation*} 9x^2+4y^2+z^2 =9x^2+4y^2+x^2+y^2 =10x^2+5y^2 =5, \end{equation*}

which we can write as \(2x^2+y^2=1\text{.}\)

The last equation describes an ellipse, so we have that \(D_0\) is given by

\(\displaystyle -\dfrac{1}{\sqrt 2}\leq x \leq\dfrac{1}{\sqrt 2}\)
\(-\sqrt{1-2x^2}\leq y\leq \sqrt{1-2x^2}\text{.}\)

From the conditions on \(z\) we finally get

\(\displaystyle \sqrt{x^2+y^2}\leq z\leq \sqrt{5-9x^2-4y^2}\)

Hence by Fubini’s Theorem we can write

\begin{equation*} \int_Df(\vect x)\,d\vect x =\int_{-1/\sqrt 2}^{1/\sqrt 2}\Biggl( \int_{-\sqrt{1-2x^2}}^{\sqrt{1-2x^2}}\Biggl( \int_{\sqrt{x^2+y^2}}^{\sqrt{5-9x^2-4y^2}} f(x,y,z)\,dz\Biggr)\,dy\Biggr)\,dx\text{.} \end{equation*}

If \(f\) is given we need to compute three integrals, starting with the innermost.