Math / Calculus

Question $\iint \sqrt{1-x^{2}-y^{2}} d x d y=\quad, x^{2}+y^{2} \leq 1$

Studdy Solution

STEP 1

1. We are given a double integral $\iint \sqrt{1-x^{2}-y^{2}} \, dx \, dy$ .
2. The region of integration is the unit disk defined by $x^2 + y^2 \leq 1$ .

STEP 2

1. Recognize the symmetry and convert to polar coordinates.
2. Set up the integral in polar coordinates.
3. Evaluate the integral with respect to $r$ .
4. Evaluate the integral with respect to $\theta$ .

STEP 3

Recognize that the region $x^2 + y^2 \leq 1$ is a circle of radius 1 centered at the origin. This suggests using polar coordinates where $x = r \cos \theta$ and $y = r \sin \theta$ .

STEP 4

Convert the integrand and differential area element to polar coordinates. The integrand becomes $\sqrt{1 - r^2}$ and the differential area element $dx \, dy$ becomes $r \, dr \, d\theta$ .
The limits for $r$ are from $0$ to $1$ , and for $\theta$ from $0$ to $2\pi$ .
The integral becomes:
$\int_{0}^{2\pi} \int_{0}^{1} \sqrt{1 - r^2} \, r \, dr \, d\theta$

STEP 5

Evaluate the inner integral with respect to $r$ :
$\int_{0}^{1} r \sqrt{1 - r^2} \, dr$
Use the substitution $u = 1 - r^2$ , $du = -2r \, dr$ , or $r \, dr = -\frac{1}{2} du$ .
Change the limits of integration: when $r = 0$ , $u = 1$ ; when $r = 1$ , $u = 0$ .
The integral becomes:
$-\frac{1}{2} \int_{1}^{0} \sqrt{u} \, du = \frac{1}{2} \int_{0}^{1} \sqrt{u} \, du$

STEP 6

Evaluate the integral $\frac{1}{2} \int_{0}^{1} \sqrt{u} \, du$ :
$\frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{0}^{1} = \frac{1}{2} \cdot \frac{2}{3} \left[ 1^{3/2} - 0^{3/2} \right] = \frac{1}{3}$

STEP 7

Evaluate the outer integral with respect to $\theta$ :
$\int_{0}^{2\pi} \frac{1}{3} \, d\theta = \frac{1}{3} \left[ \theta \right]_{0}^{2\pi} = \frac{1}{3} \cdot 2\pi = \frac{2\pi}{3}$
The value of the double integral is:
$\boxed{\frac{2\pi}{3}}$

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