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### Section 4-3 : Double Integrals over General Regions

9. Evaluate $$\displaystyle \iint\limits_{D}{{7{x^2} + 14y\,dA}}$$ where $$D$$ is the region bounded by $$x = 2{y^2}$$ and $$x = 8$$ in the order given below.

1. Integrate with respect to $$x$$ first and then $$y$$.
2. Integrate with respect to $$y$$ first and then $$x$$.
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a Integrate with respect to $$x$$ first and then $$y$$. Show All Steps Hide All Steps
Start Solution

Here’s a quick sketch of the region with the curves labeled for integration with respect to $$x$$ first.

The limits for the integral for integration with respect to $$x$$ first are then,

$\begin{array}{c} - 2 \le y \le 2\\ \,2{y^2} \le x \le 8\end{array}$

Plugging these limits into the integral is then,

$\iint\limits_{D}{{7{x^2} + 14y\,dA}} = \int_{{ - 2}}^{2}{{\int_{{2{y^2}}}^{8}{{7{x^2} + 14y\,dx}}\,dy}}$ Show Step 2

The $$x$$ integration for this integral is,

\begin{align*}\iint\limits_{D}{{7{x^2} + 14y\,dA}} & = \int_{{ - 2}}^{2}{{\int_{{2{y^2}}}^{8}{{7{x^2} + 14y\,dx}}\,dy}}\\ & = \int_{{ - 2}}^{2}{{\left. {\left( {\frac{7}{3}{x^3} + 14xy} \right)} \right|_{2{y^2}}^8\,dy}}\\ & = \int_{{ - 2}}^{2}{{\frac{{3584}}{3} + 112y - 28{y^3} - \frac{{56}}{3}{y^6}\,dy}}\end{align*} Show Step 3

Finally, the $$y$$ integration is,

\begin{align*}\iint\limits_{D}{{7{x^2} + 14y\,dA}} & = \int_{{ - 2}}^{2}{{\frac{{3584}}{3} + 112y - 28{y^3} - \frac{{56}}{3}{y^6}\,dy}}\\ & = \left. {\left( {\frac{{3584}}{3}y + 56{y^2} - 7{y^4} - \frac{8}{3}{y^7}} \right)} \right|_{ - 2}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{4096}}\end{align*}

b Integrate with respect to $$y$$ first and then $$x$$. Show All Steps Hide All Steps
Start Solution

Here’s a quick sketch of the region with the curves labeled for integration with respect to $$y$$ first.

Note that in order to do $$y$$ integration first we needed to solve the equation of the parabola for $$y$$ so the top and bottom curve will have distinct equations in terms of $$x$$, which we need to integrate with respect to $$y$$ first.

The limits for the integral for integration with respect to $$y$$ first are then,

$\begin{array}{c} \displaystyle 0 \le x \le 8\\ - \sqrt {\frac{1}{2}x} \le y \le \sqrt {\frac{1}{2}x} \end{array}$

Plugging these limits into the integral is then,

$\iint\limits_{D}{{7{x^2} + 14y\,dA}} = \int_{0}^{8}{{\int_{{ - \sqrt {\frac{1}{2}x} }}^{{\sqrt {\frac{1}{2}x} }}{{7{x^2} + 14y\,dy}}\,dx}}$ Show Step 2

The $$y$$ integration for this integral is,

\begin{align*}\iint\limits_{D}{{7{x^2} + 14y\,dA}} & = \int_{0}^{8}{{\int_{{ - \sqrt {\frac{1}{2}x} }}^{{\sqrt {\frac{1}{2}x} }}{{7{x^2} + 14y\,dy}}\,dx}}\\ & = \int_{0}^{8}{{\left. {\left( {7{x^2}y + 7{y^2}} \right)} \right|_{ - \sqrt {\frac{1}{2}x} }^{\sqrt {\frac{1}{2}x} }\,dx}}\\ & = \int_{0}^{8}{{\frac{{14}}{{\sqrt 2 }}{x^{\frac{5}{2}}}\,dx}}\end{align*} Show Step 3

Finally, the $$x$$ integration is,

$\iint\limits_{D}{{7{x^2} + 14y\,dA}} = \int_{0}^{8}{{\frac{{14}}{{\sqrt 2 }}{x^{\frac{5}{2}}}\,dx}} = \left. {\left( {\frac{4}{{\sqrt 2 }}{x^{\frac{7}{2}}}} \right)} \right|_0^8 = \require{bbox} \bbox[2pt,border:1px solid black]{{4096}}$

We got the same result as the first order of integration as we knew we would.