Now, we convert the double integral from Cartesian coordinates to polar coordinates. We use the transformations x = rcosθ, y = rsinθ, which means x²+y² = r². The differential area element dx dy becomes r dr dθ. The limits of integration for r are from 0 to ∞ (covering the entire plane), and for θ are from 0 to 2π.

Now, we convert the double integral from Cartesian coordinates to polar coordinates. We use the transformations x = rcosθ, y = rsinθ, which means x²+y² = r². The differential area element dx dy becomes r dr dθ. The limits of integration for r are from 0 to ∞ (covering the entire plane), and for θ are from 0 to 2π.

Now we substitute the result of the inner integral back into the double integral and evaluate the outer integral with respect to θ. This gives us I² = π. Taking the square root, we find I = √(π). Since e^-x² is always positive, the integral I must be positive.

Now we substitute the result of the inner integral back into the double integral and evaluate the outer integral with respect to θ. This gives us I² = π. Taking the square root, we find I = √(π). Since e^-x² is always positive, the integral I must be positive.

Prove that the Gaussian integral ∫₋∞^∞ e^(-x²) dx = √π using polar coordinates.

Answer: $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$

Step-by-step solution

Step 1: Define the integral and square it

Let the given integral be $I$ . To use polar coordinates, we need a two-dimensional integral. We can achieve this by squaring the integral. We write the second integral with respect to a different variable, $y$ , which does not change its value.

Step 2: Combine into a double integral

Since the integrals are independent, we can combine the product of two single integrals into a single double integral over the entire $xy$ -plane. The exponent becomes $-(x^2+y^2)$ .

Step 3: Convert to polar coordinates

Now, we convert the double integral from Cartesian coordinates to polar coordinates. We use the transformations $x = r\cos\theta$ , $y = r\sin\theta$ , which means $x^2+y^2 = r^2$ . The differential area element $dx dy$ becomes $r dr d\theta$ . The limits of integration for $r$ are from $0$ to $\infty$ (covering the entire plane), and for $\theta$ are from $0$ to $2\pi$ .

Step 4: Evaluate the inner integral

We evaluate the inner integral with respect to $r$ . We use a substitution $u = r^2$ , which implies $du = 2r dr$ . The limits of integration remain the same for $u$ as for $r$ . This simplifies the integral to a basic exponential form.

Step 5: Evaluate the outer integral and find I

Now we substitute the result of the inner integral back into the double integral and evaluate the outer integral with respect to $\theta$ . This gives us $I^2 = \pi$ . Taking the square root, we find $I = \sqrt{\pi}$ . Since $e^{-x^2}$ is always positive, the integral $I$ must be positive.

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