Double Integrals
figureStrips of width \(\dd x\) and \(\dd y\) under the curve \(y=f(x)\).
\[
\text{Single integral: }\int_{x_1}^{x_2}y\,\dd x,\qquad\text{double integral: }\int_{y_1}^{y_2}\!\!\int_{x_1}^{x_2}\dd x\,\dd y.
\]
\[
I_1=\int_{x_1}^{x_2}\!\!\int_{y_1}^{y_2}f(x,y)\,\dd y\,\dd x,\qquad y_1=f_1(x),\ y_2=f_2(x);\quad x_1,x_2\ \text{constant};
\]
\[
I_2=\int_{y_1}^{y_2}\!\!\int_{x_1}^{x_2}f(x,y)\,\dd x\,\dd y,\qquad x_1\to f_1(y),\ x_2\to f_2(y);\quad y_1,y_2\ \text{constant}.
\]
NoteIf \(x_1,x_2,y_1,y_2\) are all constants, either order of integration will work.
Example 6.1 Evaluate \(\displaystyle\int_0^5\!\!\int_0^{x^2}x(x^2+y^2)\,\dd y\,\dd x\).
Solution. \[
\int_0^5\Bigl[x^3y+\frac{xy^3}3\Bigr]_0^{x^2}\dd x=\int_0^5\Bigl[x^5+\frac{x^7}3\Bigr]\dd x=\Bigl[\frac{x^6}6+\frac{x^8}{24}\Bigr]_0^5=\frac{5^6}6+\frac{5^8}{24}=18880.2.
\]
Example 6.2 Evaluate \(\iint_Axy\,\dd x\,\dd y\), where \(A\) is the domain bounded by the \(x\) axis, the ordinate \(x=2a\) and the curve
\(x^2=4ay\).
Solution.
\begin{gather*}
\iint_Axy\,\dd x\,\dd y=\int_0^{2a}\!\!\int_0^{x^2/4a}xy\,\dd y\,\dd x=\int_0^{2a}\Bigl[\frac{xy^2}2\Bigr]_0^{x^2/4a}\dd x=\int_0^{2a}\frac{x^5}{32a^2}\dd x\\
=\frac16\times\frac1{32a^2}\bigl[x^6\bigr]_0^{2a}=\frac1{192a^2}\cdot2^6a^6=\frac{a^4}3.
\end{gather*}
