1.8 KiB
#uw/notes #uw/class/math224
Recall that:
\int \int_{R} f(x, y) dA
is the signed area under the function f(x,y) over the domain R where
R=[a,b] \times [c,d]
and it can be written as
\int_{a}^b \int_{c}^d f(x, y) dy dx
This can be extended to more general domains that are not rectangles such as [a,b] \times [c,d]
Example:
finding the volume under the function f(x,y) = xy over the domain bound by
\begin{gathered} 1<x<2 \\[1.5ex] x < y < x^2 \end{gathered}
We can think of finding the volume of the solid by using slices parallel to the YZ plane. With the domain provided this area can be found with
Area(x^*)=\int_{x}^{x^2}x^*ydy
Since x is bound by 1<x<2 the double integral to find the volume can be written as
\begin{gather} \int_{1}^2 \left[ \int_{x}^{x^2}xy \ dy \right] dx \\[1.5ex] \int_{1}^2 \left[ \ \left.\frac{xy^2}{2} \right\vert_{x}^{x^2} \ \right] dx \\[1.5ex] \int_{1}^2 \left[ \frac{x}{2}((x^2)^2-(x)^2) \right] dx \\[1.5ex] \int_{1}^2 \frac{x^5}{2}-\frac{x^3}{2} dx \\[1.5ex] \left.\frac{x^6}{12}-\frac{x^4}{8} \right\vert_{1}^2 \\[1.5ex] \left( \frac{16}{3} - 2 \right) - \left( \frac{1}{12} - \frac{1}{8} \right) \end{gather}
27/8
Theorem:
If the region on the XY plane for the region being integrated over of f(x,y) (D) is continuous and bound by x=a and x=b in the x-axis and two functions g_{1}(x) and g_{2}(x) on the y-axis then the integral
\int \int_{D} f(x,y)
can be rewritten as
\int_{a}^b \int_{g_{1}(x)}^{g_{2}(2)} f(x,y) \ dydx
This can also be done if D is bound by functions on the x-axis instead of the y-axis
If D=D_{1} D_{2} and D_{1} and D_{2} do not intersect except at their boundaries then
\int \int_{D} f(x,y)dA = \int \int_{D_{1}} f(x,y) \ dA + \int \int_{D_{2}} f(x,y)