Calculus III - Double Integrals over General Regions

Section 15.3 : 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.

Integrate with respect to $x$ first and then $y$.
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.

$A sketch of the region $D$ with the curves labeled for integrating with respect to $x$ first. It is bounded on the left by the parabola $x = 2{y^2}$, which opens to the right with vertex at the origin, and on the right by the vertical line $x = 8$. The inside is shaded green and it runs from $y = - 2$ to $y = 2$.$

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,

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

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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.

$A sketch of the same region $D$ with the curves labeled for integrating with respect to $y$ first. The parabola has been solved for $y$ so that the upper boundary is $y = \sqrt {\frac{1}{2}x} $ and the lower boundary is $y = - \sqrt {\frac{1}{2}x} $, with the vertical line $x = 8$ closing the region off on the right. The inside is shaded green.$

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.