Let $(x,y)$ represent the [[Cartesian coordinates|cartesian coordinates]] of a [[Parametrized Curve|curve]] in $\mathbb{R}^{2}$. The **area** between the curve $y=f(x)$ and the $x$-axis ($y=0$) in the interval $x\in(a,b)$ is the limit of the sum the area of the strips as $\Delta x\to 0$: $A= \int_{a}^{b} f(x) \, dx = \lim_{ \Delta \to 0 } \sum_{i=0}^{n-1} f(x_{i}) \Delta x $Equivalently, we can sum over area of rectangles instead of strips as follows: 1. Divide the area into tiny rectangles of area $\Delta A$. Let each rectangle have width $\Delta x=x_{i+1}-x_{i}$ and height $\Delta=y_{j+1}-y_{j}$, so that $\Delta A=\Delta x\Delta y$. 2. The area is the limit of the double sum of the area of rectangles as both $\Delta x$ and $\Delta y$ go to $0$: $A=\int_{a}^{b} \int_{0}^{f(x)} \, dy \, dx $ The **volume** between the [[Algebraic Surface|surface]] $z=f(x,y)$ and the $x$-$y$ plane $(z=0)$ can be expressed as either: 1. A double integral $\int \int f(x,y) \, dA$ (*summing up volume of thin pillars*) 2. A triple integral $\int \int \int \, dV$ (*summing up volume of tiny blocks*) # Examples > [!Example] Example 1 > Write down the area between the curve $y=x^{2}$ and $y=2x$ as a double integral and evaluate it. > >**Solution**. They intersect at $(0,0)$ and $(2,4)$. $A=\int_{0}^{2} \int_{x^{2}}^{2x} \, dy \, dx = \int_{0}^{2} 2x - x^{2} \, dx = \left[ x^{2} - \frac{x^{3}}{3} \right]^{2}_{0} =\frac{4}{3} $ > [!Example] Example 6 > Write down a triple integral representing the volume of a tetrahedron with vertices $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1)$. How many ways are there to do this? Compute the volume. > >**Solution**. $\begin{align} V&= \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \, dz \, dy \, dx \\ & = \int_{0}^{1} \int_{0}^{1-x} (1-x-y ) \, dy \, dx \\ & = \int_{0}^{1} \left[ (1-x)y-\frac{y^{2}}{2} \right]^{1-x}_{0} \, dx \\ &= \int_{0}^{1} \frac{(1-x)^{2}}{2} \, dx = \frac{1}{3} [(x-1)^{3}]^{1}_{0} = \frac{1}{6} \end{align}$ > [!Example] Example 10 (Centre of mass) > Consider a solid occupying a region $\Omega \subset \mathbb{R}^{3}$ with density $\rho(x,y,z)$ and total mass $M$. The coordinates of its [[Centre of mass (Centroid)|centre of mass]], $(\bar{x},\bar{y},\bar{z})$ are $\bar{x}=\frac{1}{M} \int \int \int_{\Omega} \rho x \, dV, \quad \bar{y} = \frac{1}{M} \int \int \int_{\Omega} \rho y \, dV, \quad \bar{z}= \frac{1}{M} \int \int \int_{\Omega} \rho z \, dV$ > Calculate the $z$-coordinate of the centre of mass of the tetrahedron in the Example 6, assuming it has uniform density $\rho=1$. > >**Solution** >Its total mass is given by $M = \int \int \int \, dx \, dy \, dz = \frac{1}{6} $Hence$\begin{align} M\bar{z} &= \int \int \int z \, dV \\ & = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} z \, dx \, dx \, dx \\ & = \frac{1}{24} \\ \implies & \bar{z} = \frac{1}{4} \end{align}$ > >Note that by symmetry, $(\bar{x},\bar{y},\bar{z}) = \left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right)$. ^116ced # Applications - [[Integration in Polar Coordinates]]. - [[Integration in cylindrical coordinates]]. - [[Integration in spherical coordinates]].