Question 9.17.3: Evaluate ∫∫R sin(x + 2y) cos (x - 2y) dA over the region R s......
Evaluate \iint_{R} \sin (x+2 y) \cos (x-2 y) d A over the region R shown in FIGURE 9.17.4(a).
The difficulty in evaluating this double integral is clearly the integrand. The presence of the terms x+2 y and x-2 y prompts us to define the change of variables u=x+2 y, v=x-2 y. These equations will map R onto a region S in the u v-plane. As in Example 1, we transform the sides of the region.
S_{1}: y=0 implies u=x and v=x or v=u. As we move from (2 \pi, 0) to (0,0), we see that the corresponding image points in the u v-plane lie on the line segment v=u from (2 \pi, 2 \pi) to (0,0).
S_{2}: x=0 implies u=2 y and v=-2 y, or v=-u. As we move from (0,0) to (0, \pi), the corresponding image points in the u v-plane lie on the line segment v=-u from (0,0) to (2 \pi,-2 \pi).
S_{3}: x+2 y=2 \pi implies u=2 \pi. As we move from (0, \pi) to (2 \pi, 0), the equation v=x-2 y shows that v ranges from v=-2 \pi to v=2 \pi. Thus, the image of S_{3} is the
vertical line segment u=2 \pi starting at (-2 \pi,-2 \pi) and going up to (2 \pi, 2 \pi). See Figure 9.17.4(b).
Now, solving for x and y in terms of u and v gives
Hence, from (11) we find that
\iint_R F(x, y) d A=\iint_S F(f(u, v), g(u, v))\left|\frac{\partial(x, y)} {\partial(u, v)}\right| d A^{\prime} . (11)
\begin{aligned} \iint\limits_{R} \sin (x+2 y) \cos (x-2 y) d A & =\iint\limits_{S} \sin u \cos v\left|-\frac{1}{4}\right| d A^{\prime} \\ & =\frac{1}{4} \int_{0}^{2 \pi} \int_{-u}^{u} \sin u \cos v d v d u \\ & \left.=\frac{1}{4} \int_{0}^{2 \pi} \sin u \sin v\right]_{-u}^{u} d u \\ & =\frac{1}{2} \int_{0}^{2 \pi} \sin ^{2} u d u \\ & =\frac{1}{4} \int_{0}^{2 \pi}(1-\cos 2 u) d u \\ & =\frac{1}{4}\left[u-\frac{1}{2} \sin 2 u\right]_{0}^{2 \pi}=\frac{\pi}{2} . \end{aligned}