Double integrals are the method to integrate over 2-D area. Double Integrals have many uses, the most popular are calculating the area of a region, the volume under a surface or the average value of a function over a plane region.
Practice Problems
\(\textbf{1)}\) \(\displaystyle\int_{{\,-3}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{x^2+y^3\,dx}}\,dy}}\)
\(\,\,\,\)The answer is \(\frac{721}{12} \approx 60.0833\)
\(\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{x^2+y^3\,dx}}\,dy}}\)
\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)
\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}}{{\left(\frac{x^3}{3}+xy^3\Big|_1^2\right)}} \,dy\)
\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}} {{\left(\frac{(2)^3}{3}+(2)y^3\right)-\left(\frac{(1)^3}{3}+(1)y^3\right)}} \,dy\)
\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}} {{\left(\frac{8}{3}+2y^3-\frac{1}{3}-y^3\right)}} \,dy\)
\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}} {{\frac{7}{3}+y^3}} \,dy\)
\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)
\(\,\,\,\,\,\,\,\,\,\displaystyle {{\frac{7}{3} \cdot y +\frac{y^4}{4}}} \Big|_{-3}^4\)
\(\,\,\,\,\,\,\,\,\,\displaystyle {{\left(\frac{7}{3} \cdot (4) +\frac{(4)^4}{4} \right)- \left(\frac{7}{3} \cdot (-3) +\frac{(-3)^4}{4}\right)}}\)
\(\,\,\,\,\,\,\,\,\,\displaystyle {{\frac{28}{3} + 64 + 7 – \frac{81}{4}}}\)
\(\,\,\,\displaystyle \frac{721}{12}\approx 60.0833\)
\(\textbf{2)}\) \(\displaystyle\int_{{\,0}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{2x\,dx}}\,dy}}\)
The answer is \(12\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{2x\,dx}}\,dy}}\)
\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{\frac{2x^2}{2} \Big|_1^2\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{x^2 \Big|_1^2\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{(2)^2-(1)^2\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{4-1\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{3\,dy}}\)
\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle{{3y\Big|_0^4}}\)
\(\,\,\,\,\,\,\displaystyle{{3(4)-3(0)}}\)
The answer is \(12\)
\(\textbf{3)}\) \(\displaystyle\int_{{\,0}}^{{\,2}}{{\int_{{\,5}}^{{\,6}}{{18x^2y^3\,dx}}\,dy}}\)
The answer is \(2{,}184\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{\int_{{\,5}}^{{\,6}}{{18x^2y^3\,dx}}\,dy}}\)
\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{\frac{18x^3y^3}{3}\,\Big|_5^6\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{6x^3y^3\,\Big|_5^6\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{6(6)^3y^3-6(5)^3y^3\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{1296y^3-750y^3\,dy}}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{546y^3\,dy}}\)
\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle{{\frac{546y^4}{4}\Big|_0^2}}\)
\(\,\,\,\,\,\,\displaystyle{{136.5y^4 \Big|_0^2}}\)
\(\,\,\,\,\,\,\displaystyle{{136.5(2)^4-136.5(0)^4}}\)
\(\,\,\,\,\,\,\displaystyle{{136.5 \cdot (16)}}\)
\(\,\,\,\,\,\,\)The answer is \(2{,}184\)
\(\textbf{4)}\) \(\displaystyle\int_{{\,\pi}}^{{\,4\pi}}{{\int_{{\,0}}^{{\,2\pi}}{{\cos x- \sin y\,dx}}\,dy}}\)
The answer is \(4 \pi\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,\pi}}^{{\,4\pi}}{{\int_{{\,0}}^{{\,2\pi}}{{\cos x- \sin y\,dx}}\,dy}}\)
\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,\pi}}^{{\,4\pi}}{{\left(\sin x-x\sin y\Big|_0^{2\pi}\right)}}\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,\pi}}^{{\,4\pi}}{{\left(\sin(2\pi)-2\pi\sin y\right)-\left(\sin(0)-0\sin y\right)}}\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,\pi}}^{{\,4\pi}}{{-2\pi\sin y}}\,dy\)
\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle 2\pi\cos y\Big|_{\pi}^{4\pi}\)
\(\,\,\,\,\,\,\displaystyle 2\pi\cos(4\pi)-2\pi\cos(\pi)\)
\(\,\,\,\,\,\,\displaystyle 2\pi(1)-2\pi(-1)\)
\(\,\,\,\,\,\,\)The answer is \(4\pi\)
\(\textbf{5)}\) \(\displaystyle\int_{{\,2}}^{{\,3}}{{\int_{{\,-1}}^{{\,1}}{{\frac{1}{(x+y)^3}\,dx}}\,dy}}\)
The answer is \(\frac{5}{24} \approx 0.20833\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,2}}^{{\,3}}{{\int_{{\,-1}}^{{\,1}}{{\frac{1}{(x+y)^3}\,dx}}\,dy}}\)
\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,2}}^{{\,3}}{{-\frac{1}{2(x+y)^2}\Big|_{-1}^{1}}}\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,2}}^{{\,3}}{{\left(-\frac{1}{2(y+1)^2}+\frac{1}{2(y-1)^2}\right)}}\,dy\)
\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\left(\frac{1}{2(y+1)}-\frac{1}{2(y-1)}\right)\Big|_2^3\)
\(\,\,\,\,\,\,\displaystyle\left(\frac{1}{8}-\frac{1}{4}\right)-\left(\frac{1}{6}-\frac{1}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle -\frac{1}{8}+\frac{1}{3}\)
\(\,\,\,\,\,\,\)The answer is \(\frac{5}{24} \approx 0.20833\)
\(\textbf{6)}\) \(\displaystyle\int_{{\,1}}^{{\,2}}{{\int_{{\,1}}^{{\,2}}{{e^{xy}\,dx}}\,dy}}\)
The answer is \(\approx 11.6175\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\int_{{\,1}}^{{\,2}}{{e^{xy}\,dx}}\,dy}}\)
\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\frac{1}{y}e^{xy}\Big|_1^2}}\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\frac{e^{2y}-e^y}{y}}}\,dy\)
\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\frac{e^{2y}-e^y}{y}}}\,dy\)
\(\,\,\,\,\,\,\text{This integral is evaluated numerically.}\)
\(\,\,\,\,\,\,\)The answer is \(\approx 11.6175\)
\(\textbf{7)}\) \(\displaystyle\int_{{\,1}}^{{\,2}}{{\int_{{\,1}}^{{\,2}}{{x \sin y – y \sin x\,dx}}\,dy}}\)
The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\int_{{\,1}}^{{\,2}}{{x \sin y – y \sin x\,dx}}\,dy}}\)
\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\left(\frac{x^2}{2}\sin y+y\cos x\Big|_1^2\right)}}\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\left(2\sin y+y\cos2\right)-\left(\frac{1}{2}\sin y+y\cos1\right)}}\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{{\,1}}^{{\,2}}{{\frac{3}{2}\sin y+y(\cos2-\cos1)}}\,dy\)
\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)
\(\,\,\,\,\,\,\displaystyle\left(-\frac{3}{2}\cos y+\frac{y^2}{2}(\cos2-\cos1)\right)\Big|_1^2\)
\(\,\,\,\,\,\,\displaystyle\frac{3}{2}(\cos1-\cos2)+\frac{3}{2}(\cos2-\cos1)\)
\(\,\,\,\,\,\,\)The answer is \(0\)
\(\textbf{8)}\) \(\displaystyle\int_{0}^{1} \int_{0}^{2} x + y \, dy \, dx\)
The answer is \(3\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{0}^{2} x + y \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left( x \cdot y + \frac{y^2}{2} \Big|_{0}^{2} \right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(2x + 2 \right) dx\)
\(\,\,\,\,\,\,\left( x^2 + 2x \Big|_{0}^{1} \right)\)
\(\,\,\,\,\,\,\)The answer is \(3\)
\(\textbf{9)}\) \(\displaystyle\int_{0}^{1} \int_{0}^{x} 2y \, dy \, dx\)
The answer is \(\frac{1}{3}\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{0}^{x} 2y \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(y^2 \Big|_{0}^{x} \right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} x^2 dx = \frac{x^3}{3} \Big|_{0}^{1}\)
\(\,\,\,\,\,\,\)The answer is \(\frac{1}{3}\)
\(\textbf{10)}\) \(\displaystyle\int_{0}^{1} \int_{1}^{2} 3x + y \, dx \, dy\)
The answer is \(5\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{1}^{2} 3x + y \, dx \, dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left( \frac{3x^2}{2} + yx \Big|_{1}^{2} \right) dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(\frac{9}{2} + y \right) dy\)
\(\,\,\,\,\,\,\left(\frac{9}{2}y + \frac{y^2}{2} \Big|_{0}^{1} \right)\)
\(\,\,\,\,\,\,= \frac{9}{2} \cdot 1 + \frac{1^2}{2} – 0 = \frac{9}{2} + \frac{1}{2} = 5\)
\(\,\,\,\,\,\,\)The answer is \(5\)
\(\textbf{11)}\) \(\displaystyle\int_{0}^{2} \int_{-1}^{1} xy \, dy \, dx\)
The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \int_{-1}^{1} xy \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \left(\frac{xy^2}{2}\Big|_{-1}^{1}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \left(\frac{x}{2}-\frac{x}{2}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} 0\,dx\)
\(\,\,\,\,\,\,\)The answer is \(0\)
\(\textbf{12)}\) \(\displaystyle\int_{0}^{2} \int_{0}^{2} x^2 + y^2 \, dy \, dx\)
The answer is \(\frac{32}{3}\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \int_{0}^{2} x^2 + y^2 \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \left(x^2y+\frac{y^3}{3}\Big|_{0}^{2}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \left(2x^2+\frac{8}{3}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\left(\frac{2x^3}{3}+\frac{8x}{3}\Big|_{0}^{2}\right)\)
\(\,\,\,\,\,\,\displaystyle \frac{16}{3}+\frac{16}{3}\)
\(\,\,\,\,\,\,\)The answer is \(\frac{32}{3}\)
\(\textbf{13)}\) \(\displaystyle\int_{-1}^{1} \int_{-1}^{1} x^2 – y^2 \, dy \, dx\)
The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{1} \int_{-1}^{1} x^2 – y^2 \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{1} \left(x^2y-\frac{y^3}{3}\Big|_{-1}^{1}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{1} \left(\frac{2x^2}{1}-\frac{2}{3}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\left(\frac{2x^3}{3}-\frac{2x}{3}\Big|_{-1}^{1}\right)\)
\(\,\,\,\,\,\,\)The answer is \(0\)
\(\textbf{14)}\) \(\displaystyle\int_{0}^{1} \int_{0}^{2} x e^{y} \, dy \, dx\)
The answer is \(\displaystyle \frac{(e^2 – 1)}{2}\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{0}^{2} x e^{y} \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(xe^y\Big|_{0}^{2}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} x(e^2-1)\,dx\)
\(\,\,\,\,\,\,\displaystyle (e^2-1)\frac{x^2}{2}\Big|_0^1\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{(e^2 – 1)}{2}\)
\(\textbf{15)}\) \(\displaystyle\int_{0}^{2} \int_{0}^{2} x \cdot y \, dy \, dx\)
The answer is \(4\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \int_{0}^{2} x \cdot y \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \left(\frac{xy^2}{2}\Big|_0^2\right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} 2x\,dx\)
\(\,\,\,\,\,\,\displaystyle x^2\Big|_0^2\)
\(\,\,\,\,\,\,\)The answer is \(4\)
\(\textbf{16)}\) \(\displaystyle\int_{0}^{3} \int_{0}^{2} x+y \, dx \, dy\)
The answer is \(15\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \int_{0}^{2} x+y \, dx \, dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left(\frac{x^2}{2}+xy\Big|_{0}^{2}\right) dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left(2+2y\right) dy\)
\(\,\,\,\,\,\,\displaystyle\left(2y+y^2\Big|_{0}^{3}\right)\)
\(\,\,\,\,\,\,\)The answer is \(15\)
\(\textbf{17)}\) \(\displaystyle\int_{0}^{1} \int_{0}^{1} 6xy \, dx \, dy\)
The answer is \(\frac{3}{2}\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{0}^{1} 6xy \, dx \, dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(3x^2y\Big|_{0}^{1}\right) dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} 3y\,dy\)
\(\,\,\,\,\,\,\displaystyle\frac{3y^2}{2}\Big|_{0}^{1}\)
\(\,\,\,\,\,\,\)The answer is \(\frac{3}{2}\)
\(\textbf{18)}\) \(\displaystyle\int_{1}^{3} \int_{0}^{1} 4x^3+2y \, dy \, dx\)
The answer is \(82\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \int_{0}^{1} 4x^3+2y \, dy \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left(4x^3y+y^2\Big|_{0}^{1}\right) dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left(4x^3+1\right) dx\)
\(\,\,\,\,\,\,\displaystyle\left(x^4+x\Big|_{1}^{3}\right)\)
\(\,\,\,\,\,\,\displaystyle(81+3)-(1+1)\)
\(\,\,\,\,\,\,\)The answer is \(82\)
\(\textbf{19)}\) \(\displaystyle\int_{0}^{2} \int_{0}^{3} y^2+2x \, dx \, dy\)
The answer is \(26\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \int_{0}^{3} y^2+2x \, dx \, dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \left(xy^2+x^2\Big|_{0}^{3}\right) dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2} \left(3y^2+9\right) dy\)
\(\,\,\,\,\,\,\displaystyle\left(y^3+9y\Big|_{0}^{2}\right)\)
\(\,\,\,\,\,\,\displaystyle8+18\)
\(\,\,\,\,\,\,\)The answer is \(26\)
\(\textbf{20)}\) \(\displaystyle\int_{0}^{\pi} \int_{0}^{1} x\cos y \, dx \, dy\)
The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi} \int_{0}^{1} x\cos y \, dx \, dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi} \left(\frac{x^2}{2}\cos y\Big|_{0}^{1}\right) dy\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi} \frac{1}{2}\cos y\,dy\)
\(\,\,\,\,\,\,\displaystyle\frac{1}{2}\sin y\Big|_{0}^{\pi}\)
\(\,\,\,\,\,\,\)The answer is \(0\)
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