Answer :
Answer:
The result of the integral is [tex]\frac{\pi}{4}(\cos{4}-\cos{1})[/tex]
Step-by-step explanation:
Polar coordinates:
In polar coordinates, we have that:
[tex]r^2 = x^2 + y^2[/tex]
[tex]\int \int_{R} dA = \int \int_{R} r dr d\theta[/tex]
In which r is related to the radius values, while [tex]\theta[/tex] is related to the angles in the trigonometric circle.
In this question:
R is the region in the first quadrant
In the first quadrant in the trigonometric circles, the angles go from 0 to [tex]\frac{\pi}{2}[/tex], which means that this are the outer limits of integration.
Between the circles with center the origin and radii 1 and 2
This means that the inner limits of integration are between 1 and 2.
The integral will be given by:
[tex]\int \int_{R} \sin{x^2+y^{2}} dA = \int_{0}^{\frac{\pi}{2}} \int_{1}^{2} \sin{r^2} r dr d\theta[/tex]
Inner integral:
[tex]\int_{1}^{2} \sin{(r^2)} r dr[/tex]
By substituion,
[tex]u = r^2[/tex]
[tex]du = 2r dr[/tex]
[tex]dr = \frac{du}{2r}[/tex]
So
[tex]\int_{1}^{2} \sin{(r^2)} r dr = \int_{1}^{2} \sin{u} r \frac{du}{2r}[/tex] = \frac{1}{2} \int_{1}^{2} \sin{u} du[/tex]
Integral of sine is minus cosine. So
[tex]\frac{1}{2}(\cos{u})|_{1}^{2}[/tex]
Before replacing, we substitute back u.
[tex]\frac{1}{2}(\cos{r^2})|_{1}^{2} = \frac{1}{2}(\cos{4}-\cos{1})[/tex]
Outer integral:
[tex]\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(\cos{4}-\cos{1}) d\theta[/tex]
[tex]\frac{\theta}{2}(\cos{4}-\cos{1})_{0}^{\frac{\pi}{2}}[/tex]
[tex]\frac{\pi}{4}(\cos{4}-\cos{1})[/tex]
The result of the integral is [tex]\frac{\pi}{4}(\cos{4}-\cos{1})[/tex]