Respuesta :
Answer:
0
Step-by-step explanation:
If we use polar coordinates, the region D can be covered by replacing (x,y) by (r*sin(Θ),rcosΘ)), with 0<r<7, 0<Θ<2π. The differential matrix
[tex]\left[\begin{array}{cc}rcos(\theta)&-rsin(\theta)\\sin(\theta)&cos(\theta)\end{array}\right][/tex]
has determinant equal to r, so we can compute the double integral as follows
[tex]\int\limits_D {xy} \, dx \, dy = \int\limits_0^{2\pi}\int\limits_0^r r^3cos(\theta)sin(\theta) \, dr \, d\theta[/tex]
(Note that we multiplied by the determinant of the Jacobian, r). A primitive for r³ is r⁴/4, thus, for Barrow's rule we have
[tex]\int\limits_0^{2\pi}\int\limits_0^r r^3cos(\theta)sin(\theta) \, dr \, d\theta = \int\limits_0^{2\pi}(\frac{r^4}{4}cos(\theta)sin(\theta)) \, |_{r = 0}^{r = 7} \, d\theta[/tex]
A primitive of cos(Θ)sin(Θ) can be obtained using substitution, and it is sin²(Θ)/2 (note that the derivate of sin²(Θ) is 2sin(Θ)cos(Θ)). Therefore, taking both the dividing 4 and the 2 obtained, we have
[tex]\int\limits_0^{2\pi}(\frac{r^4}{4}cos(\theta)sin(\theta)) \, |_{r = 0}^{r = 7} \, d\theta = \frac{1}{8} \int\limits_0^{2\pi} 7^4 * \frac{cos(\theta)sin(\theta)}{2} \, d\theta = \frac{7^4}{8} (sin^2(\theta)) |_{\theta=0}^{\theta=2\pi} \\= \frac{7^4}{8} (sin^2(2\pi)-sin^2(0)) = 0[/tex]
Hence, the integral is 0.
