Question

Evaluate the given integral by changing to polar coordinates.

Answer 1

Answer:

The result of the integral is $\frac{\pi}{4}(\cos{4}-\cos{1})$

Step-by-step explanation:

Polar coordinates:

In polar coordinates, we have that:

$r^2 = x^2 + y^2$

$\int \int_{R} dA = \int \int_{R} r dr d\theta$

In which r is related to the radius values, while $\theta$ 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 $\frac{\pi}{2}$, 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:

$\int \int_{R} \sin{x^2+y^{2}} dA = \int_{0}^{\frac{\pi}{2}} \int_{1}^{2} \sin{r^2} r dr d\theta$

Inner integral:

$\int_{1}^{2} \sin{(r^2)} r dr$

By substituion,

$u = r^2$

$du = 2r dr$

$dr = \frac{du}{2r}$

So

$\int_{1}^{2} \sin{(r^2)} r dr = \int_{1}^{2} \sin{u} r \frac{du}{2r}$ = \frac{1}{2} \int_{1}^{2} \sin{u} du[/tex]

Integral of sine is minus cosine. So

$\frac{1}{2}(\cos{u})|_{1}^{2}$

Before replacing, we substitute back u.

$\frac{1}{2}(\cos{r^2})|_{1}^{2} = \frac{1}{2}(\cos{4}-\cos{1})$

Outer integral:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(\cos{4}-\cos{1}) d\theta$

$\frac{\theta}{2}(\cos{4}-\cos{1})_{0}^{\frac{\pi}{2}}$

$\frac{\pi}{4}(\cos{4}-\cos{1})$

The result of the integral is $\frac{\pi}{4}(\cos{4}-\cos{1})$