$$
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For the function bounded within closed area \(D\), devide D into arbitrary \(n\) closed area \(\{\Delta\sigma_i | 1\le i\le n\}\), and pick arbitrary point \((\xi_i, \eta_i)\in\Delta\sigma_i\), define the double integral of function \(D\) \[ \iint_D f(x, y)\text{d}\sigma = \lim_{\lambda\rightarrow 0}\sum_{i=1}^nf(\xi_i, \eta_i)\Delta\sigma_i \] if the limit exists for all the methods of devision. The \(f(x, y)\) is called integrand, \(f(x, y)\text{d}\delta\) is the integral expression, \(\text{d}\sigma\) is the area element, \(D\) is the integral area. the sum \(\displaystyle \sum_{i=1}^nf(\xi_i, \eta_i)\Delta\sigma_i\) is named integral sum.
Specially, in the rectangular coordinates, the area element can be denoted as \(\text{d}\sigma = \text{d}x\text{d}y\), and the notion and defintion can be expressed as \[ \iint_D f(x, y)\text{d}x\text{d}y= \lim_{(x, y)\rightarrow(0,0)}\sum_{i=1}^n f(\xi_i, \eta_i)\Delta x_j\Delta y_k \] where the \(\text{d}x\text{d}y\) is called the area element in the rectangular coordinates.
Two physical significances are introduced for double integral:
The volume under curve \(f(x, y)\) \[ V = \iint f(x, y)\text{d}\sigma \]
The mass of a slice, with density distribution \(\rho(x, y)\): \[ m = \iint \rho(x, y)\text{d}\sigma \]
The property of double integral is similar to the single variable case:
1. Linearity. For the constant \(\alpha, \beta\) \[ \iint_D[\alpha f(x, y) + \beta g(x ,y )]\text{d}\sigma = \alpha \iint f(x, y)\text{d}\sigma + \beta\iint g(x, y)\text{d}\sigma \] 2. Additivity. For \(D = D_1 \cup D_2\), we have \[ \iint_D f(x, y)\text{d}\sigma = \iint_{D_1} f(x, y)\text{d}\sigma + \iint_{D_2}f(x, y)\text{d}\sigma \] 3. Keep Inequality. If \(f(x, y)\le g(x, y)\) in \(D\), then \[ \iint_D f(x, y)\text{d}\sigma \le \iint_D g(x, y)\text{d}\sigma \] Specially, notice that \(-|f(x, y)|\le f(x, y)\le |f(x, y)|\), hence \[ \left\lvert \iint_D f(x, y)\text{d}\sigma \right\rvert \le \iint_D |f(x, y)|\text{d}\sigma \] Besides, if \(M, m\) are the maximum and minimum of closed area \(D\), and \(\|D\| = \sigma'\), we have \[ m\sigma' \le \iint_D f(x, y)\text{d}\sigma \le M\sigma' \] This can be regarded as the estimation method of integral.
4. Mean Value Theorem. For the unction \(f(x, y)\) that is continous in closed area \(D\), and \(\|D\| = \sigma'\), there must be at least one point \((\xi, \eta)\) which makes \[ \iint_D f(x, y)\text{d}\sigma = f(\xi, \eta)\cdot \sigma' \] which can be easily proved by the estimation property listed in property (3).
If the integral area \(D\) is the X-type area, that is, \(D\) can be expressed by inequality \(a\le x\le b, \varphi_1(x) \le y \le \varphi_2(x)\), where the function \(\varphi_1(x), \varphi_2(x)\) is continous at interval \([a, b]\), we can prove that \[ \iint_D f(x, y)\text{d}x\text{d}y= \int_a^b \left[ \int_{\varphi_1(x)}^{\varphi_2(x)} f(x, y)\text{d}y\right] \text{d}x \] For convenience, we also denoted this expression as \[ \int_a^b \text{d}x\int_{\varphi_1(x)}^{\varphi_2(x)} f(x, y)\text{d}y \] Similarly, there also has the conversion for the Y-type area \(D\), which can be expressed by \(a\le y\le b, \varphi_1(x) \le x \le \varphi_2(x)\). \[ \iint_D f(x, y)\text{d}x\text{d}y = \int_a^b \left[ \int_{\varphi_1(y)}^{\varphi_2(y)} f(x, y)\text{d}x\right] \text{d}y = \int_a^b \text{d}y\int_{\varphi_1(y)}^{\varphi_2(y)} f(x, y)\text{d}x \] For the complex area, we can always devide it into several X-type or Y-type areas, and using the adding property of integral listed in the last node.
Several examples for the computation of double integral are provided in examples_for_double_integral_computation.
We can also convert the integral in the rectangular coordinates into polar coordinates: \[ \iint_D f(x, y)\text{d}x\text{d}y= \iint_D f(\rho\cos\theta, \rho\sin\theta)\rho~\text{d}\rho\text{d}\theta \] where the \(\rho\text{d}\rho\text{d}\theta\) is the area element.
Compute the double integral \(\displaystyle\iint_D \frac{\sin}{x}\text{d}x\text{d}y\) in \(D = \{(x, y) | 0\le x\le \pi, 0\le y\le x \}\).
This is a classical example when the sequence of integral cannot be chosen arbitrarily.
==TODO: Adding the counterpart== \[ \iint_D \frac{\sin}{x}\text{d}x\text{d}y = \int_{0}^{\pi}\text{d}x\int_{0}^{x}\frac{\sin x}{x}\text{d}y = \int_{0}^{\pi}\sin x \text{d}x = 2 \] On the contrary, if we consider it as the X-type area: \[ \iint_D \frac{\sin}{x}\text{d}x\text{d}y = \int_{0}^{\pi}\text{d}x\int_{0}^{x}\frac{\sin x}{x}\text{d}y = \int_{0}^{\pi}\sin x \text{d}x = 2 \]
Compute the double integral \(\displaystyle\iint_D x\ln(y + \sqrt{1+y^2})\text{d}x\text{d}y\) where \(D\) is constructed by the surrounding of curves \(y = 4-x^2, y = -3x, x = 1\).
This is a classical example to use the symmetry of area. \[ \begin{align} \iint_D x\ln(y + \sqrt{1+y^2})\text{d}x\text{d}y = \int_{-1}^{1}\text{d}x\int_{-3x}^{4-x^2} x\ln(y + \sqrt{1+y^2})\text{d}y \end{align} \] which is quite complex and hard to find the primitive function. The same complexity will arise if we reconsider it as the Y-type area.
We notice that \(D = D_1 \cup D_2\), where \(D_1, D_2\) are symmetric about the \(y\) and \(x\) axis respectively, combining the parity of integrand \(x\ln(y+\sqrt{1+y^2})\), we have \[ \begin{align} &\iint_D x\ln(y + \sqrt{1+y^2})\text{d}x\text{d}y\\ =& \iint_{D_1} x\ln(y + \sqrt{1+y^2})\text{d}x\text{d}y+ \iint_{D_2} x\ln(y + \sqrt{1+y^2})\text{d}x\text{d}y\\ =& 0 \end{align} \] Comment.
==TODO: Add the comment to take care of the property of the integral area itself.==
Compute \(\displaystyle\iint_D \text{e}^{-x^2-y^2}\text{d}x\text{d}y\), where \(D: x^2 + y^2 \le a^2\).
\[ \displaystyle\iint_D \text{e}^{-x^2-y^2}\text{d}x\text{d}y = \displaystyle\iint_D \text{e}^{-r^2}r\text{d}r\text{d}\theta \]
Find the volume that sphere \(x^2 + y^2 + z^2 \le 4a^2\) be cut by \(x^2+y^2 = 2ax\).
\[ V = 4\iint_D \sqrt{4a^2 - r^2}\cdot r\text{d}r\text{d}\theta \]
\[ T: J(u, v) = \frac{\partial(x, y)}{\partial(u, v)} \]