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

\(\textbf{1)}\) \(\displaystyle\int_{{\,-3}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{x^2+y^3\,dx}}\,dy}}\)

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\(\,\,\,\)The answer is \(\frac{721}{12} \approx 60.0833\)

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\(\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{x^2+y^3\,dx}}\,dy}}\)


\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)

\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}}{{\left(\frac{x^3}{3}+xy^3\Big|_1^2\right)}} \,dy\)

\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}} {{\left(\frac{(2)^3}{3}+(2)y^3\right)-\left(\frac{(1)^3}{3}+(1)y^3\right)}} \,dy\)

\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}} {{\left(\frac{8}{3}+2y^3-\frac{1}{3}-y^3\right)}} \,dy\)

\(\,\,\,\,\,\,\,\,\,\displaystyle\int_{{\,-3}}^{{\,4}} {{\frac{7}{3}+y^3}} \,dy\)


\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)

\(\,\,\,\,\,\,\,\,\,\displaystyle {{\frac{7}{3} \cdot y +\frac{y^4}{4}}} \Big|_{-3}^4\)

\(\,\,\,\,\,\,\,\,\,\displaystyle {{\left(\frac{7}{3} \cdot (4) +\frac{(4)^4}{4} \right)- \left(\frac{7}{3} \cdot (-3) +\frac{(-3)^4}{4}\right)}}\)

\(\,\,\,\,\,\,\,\,\,\displaystyle {{\frac{28}{3} + 64 + 7 – \frac{81}{4}}}\)

\(\,\,\,\displaystyle \frac{721}{12}\approx 60.0833\)

\(\textbf{2)}\) \(\displaystyle\int_{{\,0}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{2x\,dx}}\,dy}}\)

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The answer is \(12\)

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\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{2x\,dx}}\,dy}}\)



\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{\frac{2x^2}{2} \Big|_1^2\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{x^2 \Big|_1^2\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{(2)^2-(1)^2\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{4-1\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,4}}{{3\,dy}}\)



\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)

\(\,\,\,\,\,\,\displaystyle{{3y\Big|_0^4}}\)

\(\,\,\,\,\,\,\displaystyle{{3(4)-3(0)}}\)

The answer is \(12\)

\(\textbf{3)}\) \(\displaystyle\int_{{\,0}}^{{\,2}}{{\int_{{\,5}}^{{\,6}}{{18x^2y^3\,dx}}\,dy}}\)

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The answer is \(2{,}184\)

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\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{\int_{{\,5}}^{{\,6}}{{18x^2y^3\,dx}}\,dy}}\)


\(\,\,\,1^{\text{st}} \text{ step, the }1^{\text{st}} \text{ Integral}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{\frac{18x^3y^3}{3}\,\Big|_5^6\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{6x^3y^3\,\Big|_5^6\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{6(6)^3y^3-6(5)^3y^3\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{1296y^3-750y^3\,dy}}\)

\(\,\,\,\,\,\,\displaystyle\int_{{\,0}}^{{\,2}}{{546y^3\,dy}}\)


\(\,\,\,2^{\text{nd}} \text{ step, the }2^{\text{nd}} \text{ Integral}\)

\(\,\,\,\,\,\,\displaystyle{{\frac{546y^4}{4}\Big|_0^2}}\)

\(\,\,\,\,\,\,\displaystyle{{136.5y^4 \Big|_0^2}}\)

\(\,\,\,\,\,\,\displaystyle{{136.5(2)^4-136.5(0)^4}}\)

\(\,\,\,\,\,\,\displaystyle{{136.5 \cdot (16)}}\)

\(\,\,\,\,\,\,\)The answer is \(2{,}184\)

\(\textbf{4)}\) \(\displaystyle\int_{{\,\pi}}^{{\,4\pi}}{{\int_{{\,0}}^{{\,2\pi}}{{\cos x- \sin y\,dx}}\,dy}}\)

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The answer is \(4 \pi\)

\(\textbf{5)}\) \(\displaystyle\int_{{\,2}}^{{\,3}}{{\int_{{\,-1}}^{{\,1}}{{\frac{1}{(x+y)^3}\,dx}}\,dy}}\)

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The answer is \(\frac{5}{24} \approx 0.20833\)

\(\textbf{6)}\) \(\displaystyle\int_{{\,1}}^{{\,2}}{{\int_{{\,1}}^{{\,2}}{{e^{xy}\,dx}}\,dy}}\)

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The answer is \(\approx 11.6175\)

\(\textbf{7)}\) \(\displaystyle\int_{{\,1}}^{{\,2}}{{\int_{{\,1}}^{{\,2}}{{x \sin y – y \sin x\,dx}}\,dy}}\)

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The answer is \(0\)

\(\textbf{8)}\) \(\displaystyle\int_{0}^{1} \int_{0}^{2} x + y \, dy \, dx\)

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The answer is \(3\)

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\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{0}^{2} x + y \, dy \, dx\)

\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left( x \cdot y + \frac{y^2}{2} \Big|_{0}^{2} \right) dx\)

\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(2x + 2 \right) dx\)

\(\,\,\,\,\,\,\left( x^2 + 2x \Big|_{0}^{1} \right)\)

\(\,\,\,\,\,\,\)The answer is \(3\)

\(\textbf{9)}\) \(\displaystyle\int_{0}^{1} \int_{0}^{x} 2y \, dy \, dx\)

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The answer is \(\frac{1}{3}\)

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\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{0}^{x} 2y \, dy \, dx\)

\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(y^2 \Big|_{0}^{x} \right) dx\)

\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} x^2 dx = \frac{x^3}{3} \Big|_{0}^{1}\)

\(\,\,\,\,\,\,\)The answer is \(\frac{1}{3}\)

\(\textbf{10)}\) \(\displaystyle\int_{0}^{1} \int_{1}^{2} 3x + y \, dx \, dy\)

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The answer is \(5\)

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\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \int_{1}^{2} 3x + y \, dx \, dy\)

\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left( \frac{3x^2}{2} + yx \Big|_{1}^{2} \right) dy\)

\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left(\frac{9}{2} + y \right) dy\)

\(\,\,\,\,\,\,\left(\frac{9}{2}y + \frac{y^2}{2} \Big|_{0}^{1} \right)\)

\(\,\,\,\,\,\,= \frac{9}{2} \cdot 1 + \frac{1^2}{2} – 0 = \frac{9}{2} + \frac{1}{2} = 5\)

\(\,\,\,\,\,\,\)The answer is \(5\)

\(\textbf{11)}\) \(\displaystyle\int_{0}^{2} \int_{-1}^{1} xy \, dy \, dx\)

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The answer is \(0\)

\(\textbf{12)}\) \(\displaystyle\int_{0}^{2} \int_{0}^{2} x^2 + y^2 \, dy \, dx\)

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The answer is \(\frac{32}{3}\)

\(\textbf{13)}\) \(\displaystyle\int_{-1}^{1} \int_{-1}^{1} x^2 – y^2 \, dy \, dx\)

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The answer is \(0\)

\(\textbf{14)}\) \(\displaystyle\int_{0}^{1} \int_{0}^{2} x e^{y} \, dy \, dx\)

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The answer is \(\displaystyle \frac{(e^2 – 1)}{2}\)

\(\textbf{15)}\) \(\displaystyle\int_{0}^{2} \int_{0}^{2} x \cdot y \, dy \, dx\)

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The answer is \(4\)

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet \text{ Double Integral Calculator (Symbolab)}\)

\(\bullet\text{ Integration by Substitution}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int (x^2+3)^3(2x) \,dx…\)

\(\bullet\text{ Area of Region Between Two Curves}\)
\(\,\,\,\,\,\,\,\,A=\displaystyle \int_{a}^{b}\left[f(x)-g(x)\right]\,dx…\)

\(\bullet\text{ Arc Length}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}\sqrt{1+\left[f'(x)\right]^2} \,dx…\)

\(\bullet\text{ Average Function Value}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{1}{b-a} \int_{a}^{b}f(x) \,dx\)

\(\bullet\text{ Volume by Cross Sections}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Disk Method}\)
\(\,\,\,\,\,\,\,\,V=\displaystyle \int_{a}^{b}\left[f(x)\right]^2\,dx…\)

\(\bullet\text{ Cylindrical Shells}\)
\(\,\,\,\,\,\,\,\,V=2 \pi \displaystyle \int_{a}^{b} y f(y) \, dy…\)

\(\bullet\text{ Andymath Homepage}\)

In Summary

Double integrals are a mathematical concept used to calculate the area or volume of a region in the plane or in space. They are an extension of single integrals, which are used to calculate the area or length of a curve. Double integrals are important because they allow us to solve problems involving the distribution of quantities over a region. For example, we can use double integrals to calculate the mass of an object, the electric charge within a conductor, or the pressure on a surface.

Double integrals are typically introduced in a multivariate calculus course, which is typically taken by math, engineering, and physics majors in their second or third year of college. Double integrals are related to several other mathematical concepts, including triple integrals, line integrals, and surface integrals. They are also related to concepts in physics, such as mass, electric charge, and pressure.

Topics that use Double Integrals

Finding the volume of a 3D object: Double integrals can be used to find the volume of a 3D object by slicing it into infinitesimally thin horizontal or vertical slices and summing up the volumes of the slices.

Computing surface area: Double integrals can be used to find the surface area of a 3D object by slicing it into infinitesimally thin horizontal or vertical slices and summing up the areas of the slices.

Computing moments: Double integrals can be used to compute moments of a 3D object, which are used to describe the distribution of mass in the object.

Solving differential equations: Double integrals can be used to solve certain types of differential equations, which are used to describe physical systems.

Computing work: Double integrals can be used to compute the work done by a force acting on an object, which is used in mechanics and engineering.

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