Dated: 30-10-2024
Revision of Integration1
Double Integral
The double integral of a function2 of 2 variables \(x\) and \(y\) over a region \(R\) can be denoted as
\[\iint_R = f(x, y) dx dy\]
Example
Find the volume of a solid bounded by plane3 \(z = 4 - x - y\) and rectangle \(R = \{(x, y) : 0 \le x \le 1, 0 \le y \le 2\}\)
Solution
\[V = \iint_R(4 - x - y) dxdy\]
\[=\int_0^2\int_0^1 (4 - x - y) dxdy\]
\[=\int_0^2 \left[4x - \frac{x^2} 2 - xy\right]_0^1 dy\]
\[=\int_0^2 \left[4(1) - \frac{1^2} 2 - (1)y\right] - \left[4(0) - \frac{0^2} 2 - (0)y\right] dy\]
\[= \int_0^2 \left(\frac 7 2 - y\right) dy\]
\[= \left[\frac 7 2 y - \frac {y^2}{2}\right]_0^2\]
\[= \left(\frac 7 2 (2) - \frac{(2)^2} 2\right) - \left(\frac 7 2 (0) - \frac{(0)^2} 2 \right)\]
\[= 7 - 2\]
\[= 5\]
Iterated or Repeated Integral1
Following is called iterated or repeated integral
\[\int_a^b \left(\int_c^d f(x, y)dx\right) dy\]
\[\int_c^d \left(\int_a^b f(x, y)dy\right) dx\]
Here \(c \le x \le d\) and \(a \le y \le b\).
Integrating with respect to \(x\) yields a function2 of \(y\) and integrating with respect to \(y\) yields a function2 of \(x\).
Theorem
\[\int_a^b\int_c^d f(x, y) dxdy = \int_c^d\int_a^b f(x, y) dydx\]
References
Read more about notations and symbols.