Question

Evaluate the iterated integral by converting to polar coordinates. 9 0 √81 − x2 0 e−x2 − y2 dy dx

Answer 1

The question involves evaluating an iterated integral by converting to polar coordinates. The integral is over the area inside a circle in Cartesian coordinates and simplifies in polar coordinates, with the limits 0 to 9 for r and 0 to 2π for θ.

Step-by-step explanation:

The student is asking how to evaluate an iterated integral by converting from Cartesian to polar coordinates. The boundaries of integration in Cartesian coordinates appear to be the circle x2 + y2 <= 81, and the function to be integrated is e-(x2 + y2). When converting to polar coordinates, x = r*cos(θ), y = r*sin(θ), and the differential area element dxdy becomes r dr dθ. The function to be integrated simplifies to e-r2, and the new limits of integration become 0 to 2π for θ and 0 to 9 for r.

To evaluate the integral, one must integrate e-r2 with respect to r from 0 to 9, and then integrate the result with respect to θ from 0 to 2π. Adding the results from both integrations gives the final answer to the iterated integral.

Answer 2

The iterated integral is evaluated by converting to polar coordinates with limits for r from 0 to 9 and theta from 0 to \(\pi/2\). The integral simplifies to an exponential function in terms of r, and the angular integral is a straightforward integration over theta.

Step-by-step explanation:

To evaluate the iterated integral \(\int_{0}^{9} \int_{0}^{\sqrt{81 - x^2}} e^{-x^2 - y^2} dy dx\) by converting to polar coordinates, we identify the region of integration as a quarter circle with radius 9 in the first quadrant. The corresponding polar limits for r will be from 0 to 9, while the polar angle \(\theta\) varies from 0 to \(\frac{\pi}{2}\). The integral in polar coordinates becomes \(\int_{0}^{\frac{\pi}{2}} \int_{0}^{9} e^{-(r^2)} r dr d\theta\).

The factor of r comes from the conversion of dx dy to r dr d\theta. The exponential term simplifies since r^2 represents x^2 + y^2 in polar coordinates. The integral with respect to r is straightforward and typically involves a substitution u = r^2 to simplify the process. The angular integral is then simply the integration of a constant with respect to \(\theta\).