Calculus III - Double Integrals over General Regions

Section 15.3 : Double Integrals over General Regions

8. Evaluate $ \displaystyle \iint\limits_{D}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}}$ where $D$ is the region shown below.

$The region $D$ is shaded in. It is the region between the curve $x = {y^3}$ and the y-axis, running from $y = - 1$ at the bottom up to $y = 1$ at the top. Above the x-axis the region lies to the right of the y-axis and to the left of the curve, widening out to the segment from $\left( {0,1} \right)$ to $\left( {1,1} \right)$ at the top. Below the x-axis it is the mirror image through the origin, lying to the left of the y-axis and widening out to the segment from $\left( { - 1, - 1} \right)$ to $\left( {0, - 1} \right)$ at the bottom. The two halves meet at the origin.$

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First, let’s label the two sub regions in $D$ as shown below.

$A sketch of the region $D$ split into its two sub regions. The upper piece, labeled ${D_1}$, lies above the x-axis with the curve $x = {y^3}$ as its right boundary, and the lower piece, labeled ${D_2}$, lies below the x-axis with the same curve as its left boundary. The two meet at the origin and both are shaded green.$

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Despite the fact that each of the regions is bounded by the same curve we cannot get a single set of limits that will completely describe $D$. In the upper region $x = {y^3}$ is the right boundary and in the lower region $x = {y^3}$ is the left boundary.

Therefore, each region will need a separate set of limits and so we’ll need to split the integral as follows.

\[\iint\limits_{D}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}} = \iint\limits_{{{D_{\,1}}}}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}} + \iint\limits_{{{D_{\,2}}}}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}}\]

Show Step 3

Hopefully it is clear that we’ll need to integrate $x$ first with both of the integrals. So, here are the limits for each integral.

\[\begin{array}{*{20}{c}}\begin{array}{c}{D_{\,1}}\\ \,0 \le y \le 1\\ 0 \le x \le {y^3}\end{array}&{\hspace{0.5in}}&\begin{array}{c}{D_{\,2}}\\ - 1 \le y \le 0\\ {y^3} \le x \le 0\end{array}\end{array}\]

The integrals are then,

\[\iint\limits_{D}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}} = \int_{0}^{1}{{\int_{0}^{{{y^3}}}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dx}}\,dy}} + \int_{{ - 1}}^{0}{{\int_{{{y^3}}}^{0}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dx}}\,dy}}\]

Show Step 4

Not much to do now other than do the integrals. Here is the $x$ integration for both of them.

\[\begin{align*}\iint\limits_{D}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}} & = \int_{0}^{1}{{\left. {\left( {x{{\bf{e}}^{\,{y^{\,4}}}}} \right)} \right|_0^{{y^3}}\,dy}} + \int_{{ - 1}}^{0}{{\left. {\left( {x{{\bf{e}}^{\,{y^{\,4}}}}} \right)} \right|_{{y^3}}^0\,dy}}\\ & = \int_{0}^{1}{{{y^3}{{\bf{e}}^{\,{y^{\,4}}}}\,dy}} + \int_{{ - 1}}^{0}{{ - {y^3}{{\bf{e}}^{\,{y^{\,4}}}}\,dy}}\end{align*}\]

Show Step 5

Finally, here is the $y$ integration for both of the integrals.

\[\begin{align*}\iint\limits_{D}{{{{\bf{e}}^{\,{y^{\,4}}}}\,dA}} & = \int_{0}^{1}{{{y^3}{{\bf{e}}^{\,{y^{\,4}}}}\,dy}} + \int_{{ - 1}}^{0}{{ - {y^3}{{\bf{e}}^{\,{y^{\,4}}}}\,dy}}\\ & = \left. {\left( {\frac{1}{4}{{\bf{e}}^{\,{y^{\,4}}}}} \right)} \right|_0^1 + \left. {\left( { - \frac{1}{4}{{\bf{e}}^{\,{y^{\,4}}}}} \right)} \right|_{ - 1}^0\\ & = \,\frac{1}{4}\left( {{\bf{e}} - 1} \right) + \frac{1}{4}\left( { - 1 + {\bf{e}}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\left( {{\bf{e}} - 1} \right) = 0.8591}}\end{align*}\]

Don’t always expect every integral over a region to be done with a single integral. On occasion you will need to split the integral up and do the actual integration over separate sub regions. In this case that was fairly obvious but sometimes it might not be so clear until you get into the problem and realize it would be easier to do over sub regions.