Calculus III - Double Integrals over General Regions

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Section 15.3 : Double Integrals over General Regions

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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.

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.