Problem 14 - University Calculus: Early Transcendentals

Question

Convert the integral $$\int_{-1}^{1} \int_{0}^{\sqrt{1-y^{2}}} \int_{0}^{x}\left(x^{2}+y^{2}\right) d z d x d y$$ to an equivalent integral in cylindrical coordinates and evaluate the result.

Step-by-Step Solution

Answer

The value of the integral is \( \frac{2}{5} \).

Step 1: Understand the Region of Integration

The given integral is in Cartesian coordinates and integrates over the region described by \( z \) ranging from 0 to \( x \), \( x \) from 0 to \( \sqrt{1-y^2} \), and \( y \) from -1 to 1. This region suggests a part of a cylinder with respect to the \( x \)-axis, given \( x = \sqrt{1-y^2} \) forms a semicircle in the \( xy \)-plane.

Step 2: Setup Cylindrical Coordinates

Cylindrical coordinates are \( (r, heta, z) \), where \( x = r \cos \ heta, y = r \sin \ heta, z = z, \) and the Jacobian for the transformation is \( r \). The angle \( \ heta \) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), \( r \) ranges from 0 to 1, and \( z \) ranges from 0 to \( r \cos \ heta \).

Step 3: Convert the Integral

The integral becomes \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{1} \int_{0}^{r \cos \theta} \left((r \cos \theta)^2 + (r \sin \theta)^2\right) r \, dz \, dr \, d\theta \]Simplifying the expression, \( (r \cos \theta)^2 + (r \sin \theta)^2 = r^2 \). This results in \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{1} \int_{0}^{r \cos \theta} r^3 \, dz \, dr \, d\theta \]

Step 4: Evaluate the Integral with Respect to z

Integrate with respect to \( z \):\[ \int_{0}^{r \cos \theta} r^3 \, dz = r^3z \bigg|_{0}^{r\cos\theta} = r^4 \cos \theta \]

Step 5: Evaluate with Respect to r

Now integrate with respect to \( r \):\[ \int_{0}^{1} r^4 \cos \theta \, dr = \cos \theta \left[ \frac{r^5}{5} \right]_{0}^{1} = \frac{1}{5} \cos \theta \]

Step 6: Evaluate with Respect to \( \theta \)

Finally, integrate with respect to \theta:\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{5} \cos \theta \, d\theta = \frac{1}{5} \left[ \sin \theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{5} (\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})) = \frac{1}{5} (1 - (-1)) = \frac{2}{5}\]

Step 7: Compile the Result

Thus, the value of the integral evaluated in cylindrical coordinates is \( \frac{2}{5} \).

Key Concepts

Triple Integral

A triple integral is a powerful tool in calculus for finding the volume of a region or the mass of a solid object, among other things. It extends the concept of a single integral, which finds the area under a curve, or double integral, which finds the volume under a surface, into three dimensions. In this context, the triple integral \( \int_{-1}^{1} \int_{0}^{\sqrt{1-y^{2}}} \int_{0}^{x}(x^{2}+y^{2})\, dz \ dx \ dy \) is used to describe integration over a three-dimensional region in space. When solving triple integrals, the order of integration is quite important. It can define boundaries for each variable and influence the difficulty of the calculation. Typically, you will integrate patterns with respect to \( z \), then \( x \), and finally \( y \) or whatever order is prescribed by the limits given. In our example, the integral evaluates from the inside-out, starting with \( z \). Triple integrals can initially seem complex, but breaking the problem into smaller steps makes them more approachable. Start by understanding the region you are integrating over, and visualize it if possible for better comprehension.

Coordinate Transformation

A coordinate transformation, like changing Cartesian coordinates to cylindrical coordinates, is essential when solving integrals because it can simplify the integration process or make a problem solvable at all. Cylindrical coordinates make problems with symmetry around a central axis more straightforward. These coordinates are described by \( (r, \theta, z) \), where:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( z = z \)

The conversion involves understanding the geometry of the region you're dealing with, often by using trigonometric identities or visualization techniques.In this problem, we use cylindrical coordinates to convert a complex region in Cartesian coordinates into a simpler cylindrical shape, described as \( z \) running from 0 to \( r \cos \theta \), \( r \) from 0 to 1, and \( \theta \) from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). The Jacobian from this transformation, which scales the volume element \( dx \ dy \ dz \) to \( r \, dr \, d\theta \ dz \), simplifies calculations.Using cylindrical coordinates thus simplifies the triple integral by taking advantage of the symmetry of regions.

Integral Calculus

Integral calculus is the branch of calculus focused on accumulation of quantities and the areas under and between curves. To solve complex integrals, like a triple integral in cylindrical coordinates, breaking the problem into manageable parts is a useful strategy. The integration sequence aligns with the limits of integration, which are set up based on the problem's constraints. Within the cylindrical coordinates, we solve iteratively as per the axis direction.The process we followed includes:

• First integrating with respect to \( z \), yielding an expression independent of \( z \).
• Next, integrating with respect to \( r \), considering the product rule if necessary.
• Finally, integrating over \( \theta \), which often simplifies through common trigonometric identities or known integral results.

Evaluating each integral step-by-step gradually reduces the multi-layered integration into a simple numerical solution. Through careful application of integral calculus principles, we find the evaluated result as \( \frac{2}{5} \), illustrating the power and elegance of these methods in solving complex spatial problems.