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Section 15.3 : Double Integrals over General Regions

5. Evaluate \( \displaystyle \iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}}\) where \(D\) is the region bounded by \(y = 2\), \(\displaystyle y = \frac{1}{4}{x^2}\) and the \(y\)-axis.

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Start Solution

Below is a quick sketch of the region \(D\).

Note that we gave both forms of the equation for the lower curve to help with the next step.

In general, this sketch is often important to setting the integral up correctly. We’ll need to determine the order of integration and often the region will “force” a particular order. Many regions can only be dealt with easily by doing one particular order of integration and sometimes the only way to really see that is to have a sketch of \(D\).

Even if you can do the integral in either order the sketch of \(D\) will often help with setting up the limits for the integrals.

Show Step 2

With this problem the region can be dealt with either order of integration. The integral on the other hand can’t. We simply cannot integrate \(y\) first as there is no \({y^2}\) in front of the cosine we’d need to do the substitution. So, we’ll have no choice but to do the \(x\) integration first.

So, here are the limits for this integral.

\[\begin{array}{c}0 \le y \le 2\\ 0 \le x \le 2\sqrt y \end{array}\] Show Step 3

Here is the integral set up for \(x\) integration first.

\[\iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}} = \int_{0}^{2}{{\int_{0}^{{2\sqrt y }}{{5{x^3}\cos \left( {{y^3}} \right)\,dx}}\,dy}}\] Show Step 4

Here is the \(x\) integration.

\[\begin{align*}\iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}} & = \int_{0}^{2}{{\int_{0}^{{2\sqrt y }}{{5{x^3}\cos \left( {{y^3}} \right)\,dx}}\,dy}}\\ & = \int_{0}^{2}{{\left. {\left( {\frac{5}{4}{x^4}\cos \left( {{y^3}} \right)} \right)} \right|_0^{2\sqrt y }\,dy}}\\ & = \int_{0}^{2}{{20{y^2}\cos \left( {{y^3}} \right)\,dy}}\end{align*}\] Show Step 5

Note that while we couldn’t do the \(y\) integration first we can do it now. The \({y^2}\) we need for the substitution is now there after doing the \(x\) integration.

Here is the \(y\) integration.

\[\begin{align*}\iint\limits_{D}{{5{x^3}\cos \left( {{y^3}} \right)\,dA}} & = \int_{0}^{2}{{20{y^2}\cos \left( {{y^3}} \right)\,dy}}\\ & = \left. {\left( {\frac{{20}}{3}\sin \left( {{y^3}} \right)} \right)} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{20}}{3}\sin \left( 8 \right) = 6.5957}}\end{align*}\]

Remember to have your calculator set to radians when computing to decimals.