Examples

Double integral calculation

Double integral calculation

Problem: Compute ${\int }_1^2{\int }_0^3{x}^{2}y+x{e}^{y}dxdy$.

Solution: Compute the inner integral first, treating $y$ as a constant:

$$\begin{array}{rl}{\int }_1^2{\int }_0^3{x}^{2}y+x{e}^{y}dxdy& ={\int }_1^2{\left(\frac{x^3}{3}y+\frac{x^2}{2}{e}^y\right)|}_0^{3}dy\\ & ={\int }_1^{2}9y+\frac{9}{2}{e}^{y}dy\\ & ={9\frac{y^2}{2}+\frac{9}{2}{e}^y|}_1^2\\ & =\frac{27}{2}+\frac{9}{2}(e^2-e).\end{array}$$

Iterated integral, parabolic boundary

Iterated integral, parabolic boundary

Problem: Let $a(y)=y^2$ and $b(y)=4$, $[c,d]=[0,2]$, and $f(x,y)=xy$. Compute:

$${\int }_0^2{\int }_{y^2}^{4}xydxdy.$$

Solution: We have:

$$\begin{array}{rl}{\int }_0^2{\int }_{y^2}^{4}xydxdy& ={\int }_0^2{\frac{x^2}{2}y|}_{y^2}^{4}dy={\int }_0^{2}8y-\frac{y^5}{2}dy\\ & ={4y^2-\frac{y^6}{12}|}_0^2=16-\frac{16}{3}=\frac{32}{3}.\end{array}$$

Iterated integral and volume

Iterated integral and volume

Consider the function $f(x,y)=2-x-2y$, and let $z=f(x,y)$ describe a plane.
Problem: Find the volume of the region under $z=f(x,y)$ in the first octant ($x,y,z>0$).

Solution: The region is bounded in the $xy$-plane by the axes and the line $y=(2-x)/2$, since $z=0$ there. So, compute the integral:

$$\begin{array}{rl}{\int }_0^2{\int }_0^{(2-x)/2}(2-x-2y)dydx& ={\int }_0^2{(2-x)y-y^2|}_0^{(2-x)/2}dx\\ & ={\int }_0^2\frac{(2-x{)}^2}{4}dx\\ & ={-\frac{(2-x{)}^3}{12}|}_0^2=\frac{2}{3}.\end{array}$$

The meaning of the two integration steps can be explained by writing the inner integral as the area of a cross section:

$$A(x)=\frac{(2-x{)}^2}{4}.$$

The outer integral then sums these areas multiplied by infinitesimal thicknesses $dx$: ${\int }_0^{2}A(x)dx$.

Bounds for a triangular region

Bounds for a triangular region

Problem: Write two different integration bounds for the region in the figure:

Solution: On the left, we can use ${\int }_0^4{\int }_{x/2}^{2}dydx$. On the right, use ${\int }_0^2{\int }_0^{2y}dxdy$.

Changed order of integration, new boundary functions

Changed order of integration, new boundary functions

Problem: Find new bounds to change the order of integration in the following double integral:

$${\int }_{-3}^3{\int }_0^{9-x^2}f(x,y)dydx.$$

Solution: The region can be interpreted as the portion under the top of a parabola $y=9-x^2$, facing down, maximum at $(0,3)$ in the $xy$-plane. To integrate $x$ first we take the range $x\in [-\sqrt{9-y},+\sqrt{9-y}]$ and then $y\in [0,3]$. Thus:

$${\int }_0^9{\int }_{-\sqrt{9-y}}^{+\sqrt{9-y}}f(x,y)dxdy.$$