Key Concepts

• We can use a double Riemann sum to approximate the volume of a solid bounded above by a function of two variables over a rectangular region. By taking the limit, this becomes a double integral representing the volume of the solid.
• Properties of double integral are useful to simplify computation and find bounds on their values.
• We can use Fubini’s theorem to write and evaluate a double integral as an iterated integral.
• Double integrals are used to calculate the area of a region, the volume under a surface, and the average value of a function of two variables over a rectangular region.

• A general bounded region $D$ on the plane is a region that can be enclosed inside a rectangular region. We can use this idea to define a double integral over a general bounded region.
• To evaluate an iterated integral of a function over a general nonrectangular region, we sketch the region and express it as a Type I or as a Type II region or as a union of several Type I or Type II regions that overlap only on their boundaries.
• We can use double integrals to find volumes, areas, and average values of a function over general regions, similarly to calculations over rectangular regions.
• We can use Fubini’s theorem for improper integrals to evaluate some types of improper integrals.

• To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
• The area $dA$ in polar coordinates becomes $rdrd\theta .$
• Use $x=r\text{cos}\theta ,$ $y=r\text{sin}\theta ,$ and $dA=rdrd\theta$ to convert an integral in rectangular coordinates to an integral in polar coordinates.
• Use $r^2=x^2+y^2$ and $\theta ={\text{tan}}^{\mathrm{-1}}\left(\frac{y}{x}\right)$ to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed.
• To find the volume in polar coordinates bounded above by a surface $z=f\left(r,\theta \right)$ over a region on the $xy$-plane, use a double integral in polar coordinates.

• To compute a triple integral we use Fubini’s theorem, which states that if $f\left(x,y,z\right)$ is continuous on a rectangular box $B=\left[a,b\right]\times \left[c,d\right]\times \left[e,f\right],$ then
  
  $${\displaystyle \underset{B}{\iiint }f\left(x,y,z\right)dV={\displaystyle \underset{e}{\overset{f}{\int }}{\displaystyle \underset{c}{\overset{d}{\int }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x,y,z\right)dxdydz}}}}$$
  
  and is also equal to any of the other five possible orderings for the iterated triple integral.
• To compute the volume of a general solid bounded region $E$ we use the triple integral
  
  $$V\left(E\right)={\displaystyle \underset{E}{\iiint }1dV}.$$
• Interchanging the order of the iterated integrals does not change the answer. As a matter of fact, interchanging the order of integration can help simplify the computation.
• To compute the average value of a function over a general three-dimensional region, we use
  
  $$f_{\text{ave}}=\frac{1}{V\left(E\right)}{\displaystyle \underset{E}{\iiint }f\left(x,y,z\right)}dV.$$

• To evaluate a triple integral in cylindrical coordinates, use the iterated integral
  
  $${\displaystyle \underset{\theta =\alpha }{\overset{\theta =\beta }{\int }}{\displaystyle \underset{r=g_1\left(\theta \right)}{\overset{r=g_2\left(\theta \right)}{\int }}{\displaystyle \underset{z=u_1\left(r,\theta \right)}{\overset{z=u_2\left(r,\theta \right)}{\int }}f\left(r,\theta ,z\right)}}}rdzdrd\theta .$$
• To evaluate a triple integral in spherical coordinates, use the iterated integral
  
  $${\displaystyle \underset{\theta =\alpha }{\overset{\theta =\beta }{\int }}{\displaystyle \underset{\rho =g_1\left(\theta \right)}{\overset{\rho =g_2\left(\theta \right)}{\int }}{\displaystyle \underset{\phi =u_1\left(r,\theta \right)}{\overset{\phi =u_2\left(r,\theta \right)}{\int }}f\left(\rho ,\theta ,\phi \right)}}}{\rho }^2\text{sin}\phi d\phi d\rho d\theta .$$

Finding the mass, center of mass, moments, and moments of inertia in double integrals:

• For a lamina $R$ with a density function $\rho \left(x,y\right)$ at any point $\left(x,y\right)$ in the plane, the mass is $m={\displaystyle \underset{R}{\iint }\rho \left(x,y\right)dA}.$
• The moments about the $x\text{-axis}$ and $y\text{-axis}$ are
  
  $$M_x={\displaystyle \underset{R}{\iint }y\rho \left(x,y\right)}dA\text{and}{M}_y={\displaystyle \underset{R}{\iint }x\rho \left(x,y\right)}dA.$$
• The center of mass is given by $\stackrel{\text{−}}{x}=\frac{M_y}{m},\stackrel{\text{−}}{y}=\frac{M_x}{m}.$
• The center of mass becomes the centroid of the plane when the density is constant.
• The moments of inertia about the $x-\text{axis,}$ $y-\text{axis,}$ and the origin are
  
  $$I_x={\displaystyle \underset{R}{\iint }{y}^2\rho \left(x,y\right)dA,}{I}_y={\displaystyle \underset{R}{\iint }{x}^2\rho \left(x,y\right)dA,}\text{and}{I}_0=I_x+I_y={\displaystyle \underset{R}{\iint }\left(x^2+y^2\right)\rho \left(x,y\right)dA.}$$

Finding the mass, center of mass, moments, and moments of inertia in triple integrals:

• For a solid object $Q$ with a density function $\rho \left(x,y,z\right)$ at any point $\left(x,y,z\right)$ in space, the mass is $m={\displaystyle \underset{Q}{\iiint }\rho \left(x,y,z\right)dV}.$
• The moments about the $xy\text{-plane,}$ the $xz\text{-plane,}$ and the $yz\text{-plane}$ are
  
  $$M_{xy}={\displaystyle \underset{Q}{\iiint }z\rho \left(x,y,z\right)dV},M_{xz}={\displaystyle \underset{Q}{\iiint }y\rho \left(x,y,z\right)dV},M_{yz}={\displaystyle \underset{Q}{\iiint }x\rho \left(x,y,z\right)dV}.$$
• The center of mass is given by $\stackrel{\text{−}}{x}=\frac{M_{yz}}{m},\stackrel{\text{−}}{y}=\frac{M_{xz}}{m},\stackrel{\text{−}}{z}=\frac{M_{xy}}{m}.$
• The center of mass becomes the centroid of the solid when the density is constant.
• The moments of inertia about the $yz\text{-plane,}$ the $xz\text{-plane,}$ and the $xy\text{-plane}$ are
  
  $$\begin{array}{}\\ I_x={\displaystyle \underset{Q}{\iiint }\left(y^2+z^2\right)\rho \left(x,y,z\right)dV,}{I}_y={\displaystyle \underset{Q}{\iiint }\left(x^2+z^2\right)\rho \left(x,y,z\right)dV,}\hfill \\ I_z={\displaystyle \underset{Q}{\iiint }\left(x^2+y^2\right)\rho \left(x,y,z\right)dV.}\hfill \end{array}$$

• A transformation $T$ is a function that transforms a region $G$ in one plane (space) into a region $R$ in another plane (space) by a change of variables.
• A transformation $T:G\to R$ defined as $T\left(u,v\right)=\left(x,y\right)$ $\left(\text{or}T\left(u,v,w\right)=\left(x,y,z\right)\right)$ is said to be a one-to-one transformation if no two points map to the same image point.
• If $f$ is continuous on $R,$ then ${\displaystyle \underset{R}{\iint }f\left(x,y\right)}dA={\displaystyle \underset{S}{\iint }f\left(g\left(u,v\right),h\left(u,v\right)\right)}\left|\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\right|dudv.$
• If $F$ is continuous on $R,$ then
  
  $$\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iiint }F\left(x,y,z\right)}dV& ={\displaystyle \underset{G}{\iiint }F\left(g\left(u,v,w\right),h\left(u,v,w\right),k\left(u,v,w\right)\right)}\left|\frac{\partial \left(x,y,z\right)}{\partial \left(u,v,w\right)}\right|dudvdw\hfill \\ & ={\displaystyle \underset{G}{\iiint }H\left(u,v,w\right)}\left|J\left(u,v,w\right)\right|dudvdw.\hfill \end{array}$$