Iterated Integrals

You can enter and evaluate iterated integrals . If ab, f (x)≤g(x) for all x$\left[\vphantom{ a,b}\right.$a, b$\left.\vphantom{ a,b}\right]$, and k(x, y)≥ 0 for all x$\left[\vphantom{ a,b}\right.$a, b$\left.\vphantom{ a,b}\right]$ and all y$\left[\vphantom{ f(x),g(x)}\right.$f (x), g(x)$\left.\vphantom{ f(x),g(x)}\right]$ then the iterated integral

$\displaystyle \int_{{a}}^{{b}}$$\displaystyle \int_{{f(x)}}^{{g(x)}}$k(x, y) dy dx

can be interpreted as the volume of the solid bounded by
    axb  
    f (x)≤yg(x)  
    0≤zk(x, y)  


\begin{example}
Find the volume of the solid under the surface $z=1+xy$\ and ab...
...nge the \textsf{%
Axes Type} to \textsf{Boxed}.
\end{enumerate}
\end{example}

Here are a few examples of iterated integrals.

bfIntegral sfEvaluate sfEvaluate Numerically  6pt$\dint_{{0}}^{{1}}$$\dint_{{0}}^{{x}}$x2cos y dy dx = cos 1 + 2 sin 1 - 2 = .2232442755 $\dint_{{0}}^{{1}}$$\dint_{{3y}}^{{3}}$ex2 dx dy 6pt = ${\frac{{1}}{{6}}}$e9 - ${\frac{{1}}{{6}}}$ = 1350.347321 $\dint_{{0}}^{{1}}$$\dint_{{0}}^{{x}}$sin x2 dy dx 6pt = - ${\frac{{1}}{{2}}}$cos 1 + ${\frac{{1}}{{2}}}$ = .2298488471 $\dint_{{0}}^{{1}}$$\dint_{{0}}^{{1-x}}$${\dfrac{{2y}}{{x+1}}}$ dy dx = - ${\frac{{%
5}}{{2}}}$ +4 ln 2 = .2725887222    

Other multiple integrals look innocent enough but require a bit more effort to evaluate. Attempting to evaluate the double integral

$\displaystyle \int_{{0}}^{{1}}$$\displaystyle \int_{{\sqrt{y}}}^{{1}}$$\displaystyle \sqrt{{x^{3}+1}}$ dx dy

exactly leads to frustration. However, you can reverse the order of integration by looking carefully at the region of integration in the plane. dtbpF3in2.0003in0ptThis region is bounded above by y = x2 and bounded below by y = 0. The new integral is

$\displaystyle \int_{{0}}^{{1}}$$\displaystyle \int_{{0}}^{{x^{2}}}$$\displaystyle \sqrt{{x^{3}+1}}$ dy dx

This double integral can be solved by iterated integration. The inner integral is just

$\displaystyle \int_{{0}}^{{x^{2}}}$$\displaystyle \sqrt{{x^{3}+1}}$ dy = $\displaystyle \sqrt{{\left( x^{3}+1\right) }%
}$x2

You can integrate the resulting outer integral $\int_{{0}}^{{1}}$$\sqrt{{\left(
x^{3}+1\right) }}$x2 dx by applying Calculus + Change Variable, say with u2 = x3 + 1. Then choose Evaluate and Evaluate Numerically, to get


$\displaystyle \int_{{0}}^{{1}}$$\displaystyle \sqrt{{\left( x^{3}+1\right) }}$x2 dx = $\displaystyle \int_{{1}}^{{%
\sqrt{2}}}$$\displaystyle {\frac{{2}}{{3}}}$u2 du  
  = $\displaystyle {\frac{{4}}{{9}}}$$\displaystyle \sqrt{{2}}$ - $\displaystyle {\frac{{2}}{{9}}}$  
  = .4063171388