Computing double integrals

Definition: The double integral of a continuous function $f(x,y)$ over a bounded closed and squareable region $\mathrm{\Omega }$ is defined as the number$$\underset{\mathrm{\Omega }}{\iint }f(x,y)dxdy=\underset{max|\mathrm{\Delta }{x}_i|\to 0max|\mathrm{\Delta }{y}_i|\to 0}{lim}\sum _i\sum _{j}f(x_i,y_j)\mathrm{\Delta }{x}_i\mathrm{\Delta }{y}_j,$$ where $\mathrm{\Delta }{x}_i=x_{i+1}-x_i$, $\mathrm{\Delta }{y}_j=y_{j+1}-y_j$, and the sum extends over those values of $i$ and $j$ for which $(x_i,y_j)\in \mathrm{\Omega }$.

Direct computation of the double integral.

If the region $\mathrm{\Omega }$ is defined by the inequalities $$a\le x\le b,y_1(x)\le y\le y_2(x),$$ where $y_1(x)$ and $y_2(x)$ are continuous functions on the interval $[a,b]$, then the corresponding double integral can be computed using the formula $$\underset{\mathrm{\Omega }}{\iint }f(x,y)dxdy=\underset{a}{\overset{b}{\int }}dx\underset{y_1(x)}{\overset{y_2(x)}{\int }}f(x,y)dy.$$

Change of variables in a double integral.

If continuously differentiable functions $$x=x(u,v),y=y(u,v)$$ provide a one-to-one mapping of the bounded and closed region $\mathrm{\Omega }$ in the plane $Oxy$ onto the region ${\mathrm{\Omega }}^{\prime }$ in the plane $Ouv$, and the Jacobian $$I=\frac{D(x,y)}{D(u,v)}$$ maintains a constant sign in $\mathrm{\Omega }$ except perhaps on a set of measure zero, then the formula holds true:$$\underset{\mathrm{\Omega }}{\iint }f(x,y)dxdy=\underset{{\mathrm{\Omega }}^{\prime }}{\iint }f(x(u,v),y(u,v))|I|dudv.$$

In particular, for the case of transitioning to polar coordinates $r$ and $\phi$, the formula becomes:$$\underset{\mathrm{\Omega }}{\iint }f(x,y)dxdy=\underset{{\mathrm{\Omega }}^{\prime }}{\iint }f(r\cos \phi ,r\sin \phi )rdrd\phi .$$

Examples:

Compute the integrals.

1. $\underset{0}{\overset{1}{\int }}dx\underset{0}{\overset{1}{\int }}(x+y)dy.$

Solution.

$$\underset{0}{\overset{1}{\int }}dx\underset{0}{\overset{1}{\int }}(x+y)dy=\underset{0}{\overset{1}{\int }}\left({(xy+\frac{y^2}{2})|}_0^1\right)dx=$$ $$=\underset{0}{\overset{1}{\int }}(x+\frac{1}{2}-0)dx={(\frac{x^2}{2}+\frac{1}{2}x)|}_0^1=\frac{1}{2}+\frac{1}{2}-0=1.$$

Answer: 1.

2. $\underset{0}{\overset{2\pi }{\int }}d\phi \underset{0}{\overset{a}{\int }}{r}^2{\sin }^2\phi dr.$

Solution.

$$\underset{0}{\overset{2\pi }{\int }}d\phi \underset{0}{\overset{a}{\int }}{r}^2{\sin }^2\phi dr=\underset{0}{\overset{2\pi }{\int }}(\underset{0}{\overset{a}{\int }}{r}^2{\sin }^2\phi dr)d\phi =$$ $$=\underset{0}{\overset{2\pi }{\int }}({\sin }^2\phi {\frac{r^3}{3}|}_0^a)d\phi =\underset{0}{\overset{2\pi }{\int }}{\sin }^2\phi (\frac{a^3}{3}-0)d\phi =$$

$$=\frac{a^3}{3}\underset{0}{\overset{2\pi }{\int }}\frac{1-cos2\phi }{2}d\phi =\frac{a^3}{3}{(\frac{1}{2}\phi -\frac{1}{4}\sin 2\phi )|}_0^{2\pi }=$$ $$=\frac{a^3}{3}(\frac{1}{2}2\pi -\frac{1}{4}\sin 4\phi -0)=\frac{a^3\pi }{3}.$$

Answer: $\frac{a^3\pi }{3}.$

3. What sign does the integral $\underset{|x|+|y|\le 1}{\iint }\ln (x^2+y^2),dxdy$ have?

Change the order of integration in the following integrals:

4. $\underset{0}{\overset{2}{\int }}dx\underset{x}{\overset{2x}{\int }}f(x,y)dy.$

Solution:

Let's draw the region of integration:

The region of integration is bounded by the lines $y=x$, $y=2x$, and $x=2$. Note that in this region, if $y$ varies from $0$ to $2$, then the coordinate $x$ varies from the line $y=2x$ (or $x=\frac{y}{2}$) to $y=x$. If $y$ varies from $2$ to $4$, then the coordinate $x$ varies from the line $y=2x$ ($x=\frac{y}{2}$) to $x=2$. Thus,

$$\underset{0}{\overset{2}{\int }}dx\underset{x}{\overset{2x}{\int }}f(x,y)dy=\underset{0}{\overset{2}{\int }}dy\underset{\frac{y}{2}}{\overset{x}{\int }}dx+\underset{2}{\overset{4}{\int }}dy\underset{\frac{y}{2}}{\overset{2}{\int }}dx.$$

Answer: $\underset{0}{\overset{2}{\int }}dx\underset{x}{\overset{2x}{\int }}f(x,y)dy=\underset{0}{\overset{2}{\int }}dy\underset{\frac{y}{2}}{\overset{x}{\int }}dx+\underset{2}{\overset{4}{\int }}dy\underset{\frac{y}{2}}{\overset{2}{\int }}dx.$

1. $\underset{1}{\overset{2}{\int }}dx\underset{2-x}{\overset{\sqrt{2x-x^2}}{\int }}f(x,y)dy.$

2. Evaluate the integral $\underset{\mathrm{\Omega }}{\iint }x{y}^{2}dxdy$ if the region $\mathrm{\Omega }$ is bounded by the parabola $y^2=2px$ and the line $x=p/2(p>0).$

3. In the double integral $\underset{\mathrm{\Omega }}{\iint }f(x,y)dxdy$, switch to polar coordinates $r$ and $\phi$, where $x=r\cos \phi$ and $y=r\sin \phi$. Establish the limits of integration if $\mathrm{\Omega }$ is a ring $a^2\le x^2+y^2\le b^2.$

4. Switch to polar coordinates, $r$ and $\phi$, where $x=r\cos \phi$ and $y=r\sin \phi$, and establish the limits of integration in both orders in the following integral: $\underset{0}{\overset{1}{\int }}dx\underset{1-x}{\overset{\sqrt{1-x^2}}{\int }}f(x,y)dy.$

5. Assuming $r$ and $\phi$ are polar coordinates, change the order of integration in the following integrals: $\underset{-\pi /2}{\overset{\pi /2}{\int }}d\phi \underset{0}{\overset{a\cos \phi }{\int }}f(\phi ,r)dt(a>0).$

6. Switching to polar coordinates, compute the double integral $\underset{x^2+y^2\le a^2}{\iint }\sqrt{x^2+y^2}dxdy.$

7. Instead of $x$ and $y$, introduce new variables $u$ and $v$ and define the limits of integration in the following double integrals $\underset{0}{\overset{2}{\int }}dx\underset{1-x}{\overset{2-x}{\int }}f(x,y)dy$, where $u=x+y,,,v=x-y.$

8. By making appropriate variable substitutions, reduce the double integrals to single integrals:

$\underset{\mathrm{\Omega }}{\iint }f(xy)dxdy$, where the region $\mathrm{\Omega }$ is bounded by the curves $xy=1,,,xy=2,,,y=x,,,y=4x,,(x>0,y>0).$

Tags: Change of variables in a double integral, calculus, double integrals, integral, mathematical analysis