Problem 50 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral: $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x$$ Answer: The evaluated double integral is $$\frac{207}{4}.$$

Step 1: 1. Integrate with respect to y:

Integrate the given function with respect to y, treating x as a constant: $$\int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}}dy$$ To solve this, use the substitution method: Let \(u = x^2+y^2\), then \(du = 2ydy\); so \(ydy = \frac{1}{2}du\). Now the integral becomes: $$\int \sqrt{u}\frac{1}{2}du$$ Integrating with respect to u, we get: $$\frac{1}{2}\left(\frac{2}{3}u^{\frac{3}{2}}\right)+C$$ Now, substitute back \(u = x^2+y^2\), the result is: $$\frac{1}{3}(x^{2}+y^{2})^{\frac{3}{2}}+C$$ Next, apply the limits of integration for y: $$\Big[\frac{1}{3}(x^{2}+\sqrt{9-x^2}^{2})^{\frac{3}{2}}\Big]_{0}^{\sqrt{9-x^2}}$$ Evaluate the integral at the limits and subtract: $$\frac{1}{3}(x^{2}+9-x^{2})^{\frac{3}{2}}-\frac{1}{3}(x^{2})^{\frac{3}{2}}$$ Simplify the result: $$\frac{2}{3}(9)^{\frac{3}{2}}-\frac{1}{3}(x^{2})^{\frac{3}{2}}$$

Step 2: 2. Integrate with respect to x:

Now, integrate the resulting function with respect to x in the given interval: $$\int_{0}^{3}\left(\frac{2}{3}(9)^{\frac{3}{2}}-\frac{1}{3}(x^{2})^{\frac{3}{2}}\right)dx$$ Notice that the first term is a constant, so to integrate it with respect to x, just multiply it by x. For the second term, integrate the function \(\frac{1}{3}(x^{2})^{\frac{3}{2}}\) with respect to x: $$\frac{2}{3}(9)^{\frac{3}{2}}x-\frac{1}{3}\int_{0}^{3}(x^{2})^{\frac{3}{2}}dx$$ Integrating the second term: $$-\frac{1}{3}\int_{0}^{3}x^{3}dx$$ $$-\frac{1}{3}\left[\frac{1}{4}x^{4}\right]_{0}^{3}$$ $$-\frac{1}{12}(3^{4}-0)$$ Now, plug in the limits of integration for x: $$\left[\frac{2}{3}(9)^{\frac{3}{2}}(3)-\frac{1}{12}(3^{4})\right]-\left[\frac{2}{3}(9)^{\frac{3}{2}}(0)-\frac{1}{12}(0)\right]$$ Evaluate the expression: $$\frac{2}{3}(9)^{\frac{3}{2}}(3)-\frac{1}{12}(3^{4})$$

Step 3: 3. Simplify and get the final result:

Simplify the final expression to get the evaluated double integral: $$\frac{8}{3}(27)-\frac{81}{4}$$ $$\frac{216}{3}-\frac{81}{4}$$ $$72-\frac{81}{4}$$ $$\frac{207}{4}$$ So the evaluated double integral is: $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x = \frac{207}{4}$$

Key Concepts

Polar Coordinates

Understanding polar coordinates can greatly simplify solving certain types of integrals. In polar coordinates, a point in the plane is represented by its distance from the origin and the angle it makes with the positive x-axis. This is especially useful in circular or radial symmetry problems. When using polar coordinates, the double integral of a function is expressed as:

• \( \int \int f(x, y) \, dy \, dx \rightarrow \int \int f(r \cos \theta, r \sin \theta) \cdot r \, dr \, d\theta \)
• The extra \( r \) term comes from the Jacobian determinant when transforming from Cartesian coordinates.

Here, converting a region described in Cartesian form into polar form can simplify the bounds and the function itself. For a circle of radius 3, the bounds in polar coordinates would typically be \( 0 \leq r \leq 3 \) and \( 0 \leq \theta \leq 2\pi \). This makes integration more manageable by aligning with the symmetry of the shape.

Substitution Method

The substitution method is a powerful technique when dealing with integrals, particularly those involving compound functions or nested expressions. By substituting part of the integral with a new variable, we can simplify its complexity.

• For instance, in the original problem, the expression \( \sqrt{x^2 + y^2} \) was simplified by letting \( u = x^2 + y^2 \).
• Correspondingly, the differential transform is \( du = 2y \, dy \); therefore, \( y \, dy = \frac{1}{2} \, du \).

This transformation allows us to convert the integral into terms of \( u \), simplifying the integration process. A key point in solving these problems is correctly substituting back after integration. This ensures that the solution respects the original variables before evaluating at the given limits.

Integration Techniques

Working with integrals requires a good grasp of different integration techniques to tackle various forms. Some common techniques include:

• Basic Antidifferentiation: Directly integrating simple functions based on known antiderivatives.
• Substitution: Useful for recognizing a compound function within the integral and simplifying it as previously discussed.
• Integration by Parts: Applies when dealing with products of functions, using the formula \( \int u \, dv = uv - \int v \, du \).
• Breaking Down Difficult Problems: Often involves splitting the integral into manageable parts or using symmetry to simplify calculations.

In the solution provided, after handling the substitution, the resulting expression needed straightforward integration techniques. First, handling constant terms simply involves multiplying by the integral boundaries. Polynomial terms such as \( x^3 \) can be integrated using known formulas, simplifying to \( \frac{x^4}{4} \) when evaluated over its limits.Combining these approaches allows for systematic and effective solutions to seemingly complex integrals.