Double Integral

Double integration is a mathematical process used to compute the integral of a function over a two-dimensional region. It involves two successive integrations and is often used to find areas, volumes, and other quantities that extend over a plane. Formally, if f(x, y) is a function defined over a region R in the xy -plane, the double integral of f over R is denoted as:

${\iint }_{R}f(x,y)\mathrm{dA}$

where dA represents the differential area element. The double integral can be evaluated iteratively by integrating f with respect to one variable while treating the other variable as a constant, and then integrating the resulting expression with respect to the second variable.

1.0Double Integral

An integral in which the integrand is integrated twice is a double integral,

$\iint \mathrm{f}(\mathrm{x})(\mathrm{dx}{)}^2$

The integration is performed from the inside out.

Example 1: Find ${\int }_{-1}^3{\int }_0^{2}2x(dx{)}^2.$

Solution: $\int 2xdx=x^2+C_1;{|}_0^2{x}^2=4$

$\int 4\mathrm{dx}=4\mathrm{x}+{\mathrm{C}}_2;{|}_{-1}^{3}4\mathrm{x}=16$

Hence, ${\int }_{-1}^3{\int }_0^{2}2xd{x}^2=16$

Functions of more than one variable may be integrated with respect to one variable at a time while the other variables are held constant, reversing the process of partial differentiation.

Example 2: Find ${\int }_0^3{\int }_1^2\left(4y^3+2x\right)dxdy$

Solution: Hold y constant and integrate with respect to x:

∫ (4y³ + 2x) dx = 4y³ x + x² + C;

${\left[4y^{3}x+x^2\right]}_{x=1}^{x=2}$ = (4y³. 2 + 2²) – (4y³+1²) = 4y³+ 3

Integrate with respect to y:

∫(4y³+3) dy = y⁴ + 3y + C;

${\left[y^4+3y\right]}_0^3=81+9=90.$

Hence,

${\int }_0^3{\int }_1^2\left(4y^3+2x\right)dxdy=90$

Example 3: Find  ${\int }_1^3{\int }_0^{2y}3x^{2}dxdy$.

Solution: ${\int }_1^3{\int }_0^{2y}3x^{2}dxdy={\int }_1^3{\left[x^3\right]}_{x=0}^{x=2y}dy={\int }_1^{3}8y^{3}dy={|}_1^{3}2y^4=160$

2.0Geometrical Interpretation of the Double Integral

We are familiar with the geometrical interpretation of the equation y = f(x) as a curve in the two-dimensional x, y- plane, and of the integral ${\int }_a^{b}f(x)dx$ as an area between the curve and the x- axis.

Similarly, while the equation z = f(x, y) defines a surface in the three-dimensional x, y, z - space , the double integral of a continuous function of two variables ${\int }_a^b{\int }_c^{d}f(x,y)dxdy$, may be interpreted as a volume between the surface z= f(x, y) and the x, y-plane.

In the graph, the rectangular area ΔA_i in the x, y-plane is projected on the surface z = f (x, y).

The quantity ΔA_i = Δx_i Δy_i is the area of the bottom surface of a column whose top surface is part of the surface f (x_i, yi).

The smaller the area ΔA_i, the closer the volume of the column is to that of a parallelepiped measuring f(x_i, y_i)  ΔA_i.

If the domain D, consisting of (x, y) with c ≤ x ≤ d, a ≤ y ≤ b, is divided into an increasingly greater number of rectangles so that ΔA_i tends to 0, then the volume between the surface z = f(x, y) and D equals then the sum of all parallelepipeds measuring f(x_i, y_i) Δx_i Δy_i ; thus,

$V={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f\left(x_i,y_i\right)\Delta x_i\Delta y_i={\int }_a^b{\int }_c^{d}f(x,y)dxdy$ .

3.0Solved Examples on Double Integral

Example 1: Evaluate the double integral ${\iint }_R\left(x^2+y^2\right)dA$ over the region R where $0\le x\le 1$ and $0\le y\le 1$:

Step 1: Set Up the Double Integral

${\iint }_R\left(x^2+y^2\right)dA={\int }_0^1\left({\int }_0^1\left(x^2+y^2\right)dy\right)dx$

Step 2: Integrate with Respect to y 

${\int }_0^1\left(x^2+y^2\right)dy={\left[x^{2}y+\frac{y^3}{3}\right]}_0^1=x^2(1)+\frac{1^3}{3}-0=x^2+\frac{1}{3}$

Step 3: Integrate with Respect to x 

${\int }_0^1\left(x^2+\frac{1}{3}\right)dx={\left[\frac{x^3}{3}+\frac{x}{3}\right]}_0^1=\left(\frac{1}{3}+\frac{1}{3}\right)-0=\frac{2}{3}$

Thus, the value of the double integral is $\frac{2}{3}$.

Example 2: Evaluate the double integral ${\iint }_{R}x{e}^{xy}dA$ over the region R where $0\le x\le 1$ and $0\le y\le 1$:

Step 1: Set Up the Double Integral

${\iint }_{R}x{e}^{xy}dA={\int }_0^1\left({\int }_0^{1}x{e}^{xy}dy\right)dx$

Step 2: Integrate with Respect to y 

${\int }_0^{1}x{e}^{xy}dy=x{\int }_0^1{e}^{xy}dy=x{\left[\frac{e^{xy}}{x}\right]}_0^1=x\left(\frac{e^x-1}{x}\right)=e^x-1$

Step 3: Integrate with Respect to x 

${\int }_0^1\left(e^x-1\right)dx={\left[e^x-x\right]}_0^1=\left(e^1-1\right)-(1-0)=e-2$

Thus, the value of the double integral is e – 2.

Example 3: Evaluate the double integral ${\iint }_R(3x+4y)dA$ over the triangular region R with vertices at (0,0), (1,0), and (0,1):

Step 1: Set Up the Double Integral

The region R is bounded by y = 0, x = 0, and y = 1 – x . Therefore, we integrate with respect to y first:

${\iint }_R(3x+4y)dA={\int }_0^1\left({\int }_0^{1-x}(3x+4y)dy\right)dx$

Step 2: Integrate with Respect to y

${\int }_0^{1-x}(3x+4y)dy={\left[3xy+2y^2\right]}_0^{1-x}$

=3 x(1-x)+2(1-x)^2-0

=3 x-3 x^2+2-4 x+2 x^2

=-x^2-x+2

Step 3: Integrate with Respect to x 

${\int }_0^1\left(-x^2-x+2\right)dx={\left[\frac{-x^3}{3}-\frac{x^2}{2}+2x\right]}_0^1$

$=\left(\frac{-1}{3}-\frac{1}{2}+2\right)-0$

$=\frac{7}{6}$

Thus, the value of the double integral is $\frac{7}{6}$.

Example 4: 

Evaluate the double integral ${\iint }_{R}xydA$ over the region R where $0\le x\le 2$ and $0\le y\le x$ :

Step 1: Set Up the Double Integral

${\iint }_{R}xydA={\int }_0^2\left({\int }_0^{x}xydy\right)dx$

Step 2: Integrate with Respect to y 

${\int }_0^{x}xydy=x{\int }_0^{x}ydy=x{\left[\frac{y^2}{2}\right]}_0^x=x\left(\frac{x^2}{2}-0\right)=\frac{x^3}{2}$

Step 3: Integrate with Respect to x 

${\int }_0^2\frac{x^3}{2}dx=\frac{1}{2}{\int }_0^2{x}^{3}dx=\frac{1}{2}{\left[\frac{x^4}{4}\right]}_0^2=\frac{1}{2}\left(\frac{2^4}{4}-0\right)=\frac{1}{2}\cdot \frac{16}{4}=\frac{8}{4}=2$

Thus, the value of the double integral is 2.

Example 5: Evaluate the double integral ${\iint }_R\sin (x+y)dA$ over the region R where $0\le x\le \frac{\pi }{2}$ and

$0\le y\le \frac{\pi }{2}$.

Step 1: Set Up the Double Integral

${\iint }_R\sin (x+y)dA={\int }_0^{\frac{\pi }{2}}\left({\int }_0^{\frac{\pi }{2}}\sin (x+y)dy\right)dx$

Step 2: Integrate with Respect to y 

${\int }_0^{\frac{\pi }{2}}\sin (x+y)dy=[-\cos (x+y){]}_0^{\frac{\pi }{2}}=-\cos \left(x+\frac{\pi }{2}\right)+\cos (x)=\sin (x)+\cos (x)$

Step 3: Integrate with Respect to x 

${\int }_0^{\frac{\pi }{2}}(\sin (x)+\cos (x))dx$

$=[-\cos (x)+\sin (x){]}_0^{\frac{\pi }{2}}$

$=\left(-\cos \left(\frac{\pi }{2}\right)+\sin \left(\frac{\pi }{2}\right)\right)-(-\cos (0)+\sin (0))$

=(0+1)-(-1+0)

=1+1=2

Thus, the value of the double integral is 2.

4.0Practice Questions on Double Integral

Question 1: Evaluate the double integral ${\iint }_R\left(3x^2+2y\right)dA$ over the region R where $0\le x\le 2$ and $0\le y\le 1$.

Question 2: Evaluate the double integral ${\iint }_R{e}^{x^2-y^2}dA$ over the region R where $0\le x\le 1$ and $0\le y\le 1$.

Question 3: Evaluate the double integral ${\iint }_R(x+y)dA$ over the triangular region R with vertices at (0,0), (1,0), and (1,1).

Question 4: Evaluate the double integral ${\iint }_R\cos (xy)dA$ over the region R where $0\le x\le \pi$ and $0\le y\le \pi$.

Question 5: Evaluate the double integral ${\iint }_R\frac{1}{1+x^2+y^2}dA$ over the region R where $0\le x\le 1$ and $0\le y\le 1$.

Question 6: Evaluate the double integral ${\iint }_{R}x{e}^{xy}dA$ over the region R where $0\le x\le 1$ and $0\le y\le x$.

Question 7: Evaluate the double integral ${\iint }_R\sin (x+y)dA$ over the region R where $0\le x\le \frac{\pi }{2}$ and $0\le y\le \frac{\pi }{2}$.

Question 8: Evaluate the double integral ${\iint }_R{x}^{2}ydA$ over the region R where $0\le x\le 2$ and $0\le y\le 3$ .

Question 9: Evaluate the double integral ${\iint }_R\ln (xy)dA$ over the region R where $1\le x\le 2$ and $1\le y\le 2$.

Question 10: Evaluate the double integral ${\iint }_R\frac{x^2+y^2}{1+x^2{y}^2}dA$ over the region R where $0\le x\le 1$ and $0\le y\le 1$.