Section 15.3 : Double Integrals over General Regions
9. Evaluate ∬ 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.
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 2The 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 3Finally, 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 2The 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 3Finally, 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.