Respuesta :
In polar coordinates, the region [tex]R[/tex] is the set of points
[tex]\left\{(r,\theta)\mid3\le r\le5,\,0\le\theta\le\dfrac\pi2\right\}[/tex]
and we have [tex]x^2+y^2=r^2[/tex] and [tex]\mathrm dA=r\,\mathrm dr\,\mathrm d\theta[/tex]. So the integral, converted to polar, is
[tex]\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_3^5r\sin r^2\,\mathrm dr\,\mathrm d\theta=\frac\pi2\int_3^5r\sin r^2\,\mathrm dr[/tex]
Substitute [tex]s=r^2[/tex] to get
[tex]=\displaystyle\frac\pi4\int_9^{25}\sin s\,\mathrm ds=\frac{(\cos9-\cos25)\pi}4[/tex]
Using polar coordinates, the result of the integral is: [tex]O = -0.4756\pi[/tex].
The most used relation for polar coordinates is:
[tex]x^2 + y^2 = r^2[/tex]
An integral in Cartesian coordinates is converted to polar using:
[tex]\int \int f(x,y) dA = \int \int f(r,\theta) rdrd\theta[/tex]
In this problem:
[tex]f(x,y) = \sin{x^2 + y^2}[/tex]
Thus:
[tex]f(r, \theta) = \sin{r^2}[/tex]
- Region between the circles with center at the origin and radii 3 and 5, thus the inner limits of integration are [tex]r = 3[/tex] and [tex]r = 5[/tex].
- First quadrant, thus, the outer limits of integration are: [tex]\theta = 0, \theta = \frac{\pi}{2}[/tex].
The polar integral is:
[tex]\int_{0}^{\frac{\pi}{2}} \int_{3}^5 \sin{r^2} r dr d\theta[/tex]
The inner integral is:
[tex]I = \int_{3}^5 \sin{r^2} r dr[/tex]
Using substitution:
[tex]u = r^2 \rightarrow du = 2rdr \rightarrow dr = \frac{du}{2r}[/tex]
[tex]I = \frac{1}{2} \int \sin{u} du[/tex]
[tex]I = -\frac{\cos{u}}{2}[/tex]
[tex]I = -\frac{\cos{r^2}}{2}_{3}^{5}[/tex]
Applying the fundamental theorem of calculus:
[tex]I = -\frac{\cos{5^2}}{2} + \frac{\cos{3^2}}{2} = -0.9512[/tex]
Then, the outer integral is:
[tex]O = \int_{0}^{\frac{\pi}{2}} -0.9512 d\theta[/tex]
[tex]O = -0.9512\theta_{0}^{\frac{\pi}{2}}[/tex]
[tex]O = -0.4756\pi[/tex]
The result of the integral is [tex]O = -0.4756\pi[/tex].
A similar problem is given at https://brainly.com/question/23412154
