Question

evaluate the given integral by changing to polar coordinates. e−x2 − y2 da d , where d is the region bounded by the semicircle x = 25 − y2 and the y−axis

Answer 1

The given integral
`\( \int\int_D e^(-x^2 - y^2) \, dA \)` , where D is the region bounded by the semicircle
`x = 25 - y^2 \)` and the y-axis, can be evaluated in polar coordinates as
`\( \int_(-√(25))^(√(25))\int_(0)^(√(25-y^2)) e^(-r^2) \, r \, dr \, dy \).`

Step-by-step explanation:

To evaluate the given integral by changing to polar coordinates, we express the Cartesian coordinates x and y in terms of polar coordinates r and
`\( \theta \)` . The region D is defined by the semicircle
`x = 25 - y^2 \)` and the y-axis. In polar coordinates, the semicircle equation becomes
`\( r = 25 - y^2 \), and \( \theta \)` ranges from 0 to
`\( \pi \).`

The differential area element dA in polar coordinates is
`\( r \, dr \, d\theta \)` . Substituting this into the integral, we obtain
`\( \int_(-√(25))^(√(25))\int_(0)^(√(25-y^2)) e^(-r^2) \, r \, dr \, dy \)`. This double integral represents the region D where the function
`\( e^(-x^2 - y^2) \)` is integrated over the given semicircular region.

The integral is then evaluated by integrating with respect to r from 0 to
`\( √(25-y^2) \)` and with respect to y from
`\( -√(25) \)` to
`\( √(25) \)`. This involves standard calculus techniques, and the final result gives the value of the given integral over the specified region. The use of polar coordinates simplifies the calculations, especially when dealing with circular or symmetric regions

Answer 2

To evaluate the given integral by changing to polar coordinates, convert the given equation to polar coordinates using the substitution x = rcos(theta) and y = rsin(theta). Determine the range of integration in polar coordinates by solving for r in the equation. The integral can then be expressed in polar coordinates and evaluated over the range of r and theta.

Step-by-step explanation:

To evaluate the given integral by changing to polar coordinates, we need to first determine the range of integration in polar coordinates.

The semicircle x = 25 - y^2 can be converted to polar coordinates using the substitution x = rcos(theta) and y = rsin(theta).

Substituting x and y into the equation x = 25 - y^2, we get rcos(theta) = 25 - (rsin(theta))^2.

Simplifying the equation, we have r^2cos^2(theta) = 25 - r^2sin^2(theta).

Factoring out r^2, we get r^2(cos^2(theta) + sin^2(theta)) = 25.

Since cos^2(theta) + sin^2(theta) = 1, we have r^2 = 25. Taking the square root of both sides, we get r = 5.

Therefore, the range of integration in polar coordinates is 0 <= r <= 5 and 0 <= theta <= pi.

The integral e^(-x^2 - y^2) da d can be expressed in polar coordinates as e^(-r^2) r dr d(theta).

Integrating over the range of r and theta, we have the final result:

∫(0 to pi) ∫(0 to 5) e^(-r^2) r dr d(theta).