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Sagot :
Answer:
The result of the integral is [tex]\frac{\pi}{4}(-\cos{16} + \frac{9})[/tex]
Step-by-step explanation:
Polar coordinates:
In polar coordinates, we have that:
[tex]x^2 + y^2 = r^2[/tex]
And
[tex]\int \int_{dA} f(x,y) da = \int \int f(r) r dr d\theta[/tex]
In this question:
[tex]\int \int_{dA} \sin{(x^2+y^2)} dA = \int \int_{dR} = \sin{r^2}r dr d\theta[/tex]
Region in the first quadrant between the circles with center the origin and radii 3 and 4
First quadrant means that [tex]\theta[/tex] ranges between [tex]0[/tex] and [tex]\frac{\pi}{2}[/tex]
Between these circles means that r ranges between 3 and 4. So
[tex]\int \int_{dR} = \sin{r^2}r dr d\theta = \int_{0}^{\frac{\pi}{2}} \int_{3}^{4} \sin{r^2} r dr d\theta[/tex]
Applying the inner integral:
[tex]\int_{3}^{4} \sin{r^2} r dr[/tex]
Using substitution, with [tex]u = r^2, du = 2rdr, dr = \frac{du}{2r}[/tex], and considering that the integral of the sine is minus cosine, we have:
[tex]-\frac{\cos{r^2}}{2}|_{3}{4} = \frac{1}{2}(-\cos{16} + \frac{9})[/tex]
Applying the outer integral:
[tex] \int_{0}^{\frac{\pi}{2}} \frac{1}{2}(-\cos{16} + \frac{9}) d\theta[/tex]
Has no factors of [tex]\theta[/tex], so the result is the constant multiplied by [tex]\theta[/tex], and then we apply the fundamental theorem.
[tex]\frac{\theta}{2}(-\cos{16} + \frac{9}) = \frac{\pi}{4}(-\cos{16} + \frac{9})[/tex]
The result of the integral is [tex]\frac{\pi}{4}(-\cos{16} + \frac{9})[/tex]
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