Question 27.13: Evaluate the double integral of f (x, y) = x² + 3xy over the......
Evaluate the double integral of f (x, y) = x² + 3xy over the region R indicated in Figure 27.13 by (a) integrating first with respect to x, and then with respect to y, (b) integrating first with respect to y, and then with respect to x.
(a) If we integrate first with respect to x we must select an arbitrary horizontal strip as shown in Figure 27.13 and integrate in the x direction. On OB, y = x so that the lower limit of the x integration is x = y. On AB, x = 1 so the upper limit is x = 1.
Therefore
As y varies from 0 to 1 the horizontal strips will cover the entire region. Hence the limits of integration of the y integral are 0 and 1. So
\int_{y=0}^{y=1}{\frac{1}{3}}+{\frac{3y}{2}}-{\frac{11}{6}}y^{3}\,\mathrm{d}y=\left[{\frac{1}{3}}y+{\frac{3}{4}}y^{2}-{\frac{11y^{4}}{24}}\right]_{0}^{1}\\ ={\frac{1}{3}}+{\frac{3}{4}}-{\frac{11}{24}}\\ ={\frac{15}{24}}\\ ={\frac{5}{8}}that is,
\int_{y=0}^{y=1}\int_{x=y}^{x=1}x^{2}+3x y\,\mathrm{d}x\,\mathrm{d}y={\frac{5}{8}}(b) If we choose to integrate with respect to y first we must select an arbitrary vertical strip as shown in Figure 27.14. At the lower end of the strip y = 0. At the upper end y = x.
To integrate along the strip we evaluate
We add contributions of all such vertical strips by integrating with respect to x from x = 0 to x = 1:
\int_0^1{\frac{5x^3}{2}dx} = \left[\frac{5x^4}{8}\right]_0^1 = \frac{5}{8}We see that the prudent selection of the order of integration can yield substantial savings in the effort required.