Question

Evaluate the given integral by changing to polar coordinates. sin(x2 y2) dAR, where R is the region in the first quadrant between the circles with center the origin and radii 3 and 4

Answer 1

Answer:

The result of the integral is $\frac{\pi}{4}(-\cos{16} + \frac{9})$

Step-by-step explanation:

Polar coordinates:

In polar coordinates, we have that:

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

And

$\int \int_{dA} f(x,y) da = \int \int f(r) r dr d\theta$

In this question:

$\int \int_{dA} \sin{(x^2+y^2)} dA = \int \int_{dR} = \sin{r^2}r dr d\theta$

Region in the first quadrant between the circles with center the origin and radii 3 and 4

First quadrant means that $\theta$ ranges between $0$ and $\frac{\pi}{2}$

Between these circles means that r ranges between 3 and 4. So

$\int \int_{dR} = \sin{r^2}r dr d\theta = \int_{0}^{\frac{\pi}{2}} \int_{3}^{4} \sin{r^2} r dr d\theta$

Applying the inner integral:

$\int_{3}^{4} \sin{r^2} r dr$

Using substitution, with $u = r^2, du = 2rdr, dr = \frac{du}{2r}$, and considering that the integral of the sine is minus cosine, we have:

$-\frac{\cos{r^2}}{2}|_{3}{4} = \frac{1}{2}(-\cos{16} + \frac{9})$

Applying the outer integral:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(-\cos{16} + \frac{9}) d\theta$

Has no factors of $\theta$, so the result is the constant multiplied by $\theta$, and then we apply the fundamental theorem.

$\frac{\theta}{2}(-\cos{16} + \frac{9}) = \frac{\pi}{4}(-\cos{16} + \frac{9})$

The result of the integral is $\frac{\pi}{4}(-\cos{16} + \frac{9})$