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Evaluation of a triple integral.

The method we saw in  1.2 for evaluating a double integral can be easily generalized here, together with Fubini's theorems. We will see some examples.

Example 7.12       


\begin{align*}I &= \int_0^1 \int_0^2 \int_0^1 x+yz \; dx \; dy \; dz \\
\quad &...
... \int_0^1 \frac 12 \cdot 2 \; dz \\
\quad & = \int_0^1 1 \; dz = 1
\end{align*}

Example 7.13   Compute the volume of the surface enclosed by the paraboloid whose equation is z=x2+y2 and the plane whose equation is z=1.

The two surfaces intersect on a circle whose equations are $\begin{cases}x^2+y^2=1 \\ z=1 \end{cases}$.


  
Figure 9: A cut paraboloid.
\begin{figure}
\mbox{\epsfig{file=ParaboloidSolid.eps,height=6cm} }
\end{figure}


\begin{align*}I &= \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2...
... y-x^2y - \frac 13 y^3 \right]_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}
\; dx
\end{align*}
From this point, it will be easier to use polar coordinates. We get:
\begin{align*}I &= = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (1-x^2-y^2)...
...= \frac 14 \int _0^{2 \pi} \; d \theta\\
\quad & = \frac {\pi}{2}.
\end{align*}
In fact this computation has been made using cylindrical coordinates (v.i.  2.5).



Noah Dana-Picard
2001-05-30