Question

Evaluate the iterated integral by converting to polar coordinates. $$ \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}}\left(x^{2}+y^{2}\right)^{3 / 2} d y d x $$

Short answer

The integral evaluates to \(\frac{243π}{10}\).

Step-by-step solution

Convert to polar coordinates

The limits of the inner integral go from \(y = 0\) which is \(r = 0\) up to \(y = \sqrt{9 - x^2}\) which is \(r = 3\). The limits of the outer integral go from \(x = 0\), which is \(θ = 0\), to \(x = 3\), which is \(θ = \frac{π}{2}\). Hence, the integral becomes \(\int_{0}^{π/2} \int_{0}^{3} r^3 r dr dθ = \int_{0}^{π/2} \int_{0}^{3} r^4 dr dθ\).

Evaluate the inner integral

Evaluate the integral \(\int_{0}^{3} r^4 dr\). Using the power rule for integration, the result is \(\frac{1}{5}r^5\Big|_0^3 = \frac{243}{5}\).

Evaluate the outer integral

Then replace the inner integral result into the outer integral \(\int_{0}^{π/2} \frac{243}{5}dθ\). Evaluating this results in \(\frac{243}{5}θ\Big|_0^{π/2} = \frac{243}{5}*\frac{π}{2} = \frac{243π}{10}\).

Key concepts

Polar Coordinates Conversion

Converting between Cartesian and polar coordinates is essential for simplifying complex multivariable integrals, especially when dealing with circular or spherical regions. In our exercise, the region of integration is a quarter-circle in the first quadrant, bounded by the positive x-axis, positive y-axis, and the circle with radius 3. In Cartesian coordinates, the variables x and y describe positions on a plane via horizontal and vertical distances. However, polar coordinates use a combination of the distance from the origin (r) and the angle (θ) from the positive x-axis to describe a point.

To perform the conversion, we substitute x with r*cos(θ) and y with r*sin(θ), and add a factor of r to account for the Jacobian of the transformation (the area element in polar coordinates is r*dr*dθ instead of dx*dy). Thus, the Cartesian equation y = sqrt(9 - x^2) for the quarter-circle boundary translates to the simpler polar form r = 3. With this transformation, we can set up more manageable limits of integration: from 0 to 3 for r and from 0 to π/2 for θ, covering the entire quarter-circle.

Power Rule for Integration

The power rule is a fundamental concept in calculus, providing an efficient way to integrate monomial expressions (terms like x^n). It states that the integral of x^n with respect to x is (1/(n+1))x^(n+1), plus a constant of integration for indefinite integrals. When this rule is applied to polar coordinates, the 'x' is replaced with the radial distance, 'r'.

In our problem, we have the integrand \(r^2 + y^2)^{\frac{3}{2}}\), where by converting to r terms using polar coordinates, it simplifies to \(r^4)\). Applying the power rule, the inner integral evaluates from \(r^0\) to \(r^3\) as \(\frac{3+1}{3+2}r^{3+1}\) which, after plugging in the limits of integration, simplifies to \(\frac{243}{5}\). This demonstrates how the power rule greatly simplifies integration, making complex regions much more manageable when expressed in polar form.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with more than one variable. This field encompasses the study of limits, derivatives, and integrals for multivariate functions. A key operation in multivariable calculus is the integration over domains in two or more dimensions, often requiring iterated integrals, as seen in our exercise.

An iterated integral allows for integration within a multi-dimensional region, solving one integral at a time. When evaluating such integrals, especially over circular or spherical regions, a conversion to polar coordinates simplifies the process due to the symmetry of the shape. The use of these coordinates exemplifies how the choice of the coordinate system can make a tremendous difference in the ease of computation for multivariable integrals.