Problem 16 - Calculus Early Transcendentals

Question

Evaluate the following integrals in cylindrical coordinates. The figures illustrate the region of integration. $$\int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}} \int_{0}^{9-3 \sqrt{x^{2}+y^{2}}} d z d x d y$$

Step-by-Step Solution

Answer

Answer: The value of the original triple integral is $\frac{729}{8} \cdot \pi$.

Step 1: Convert to cylindrical coordinates

First, we need to convert the integral from Cartesian coordinates to cylindrical coordinates. Recall that the transformation from Cartesian to cylindrical coordinates is given by: $$x = r \cos\theta$$ $$y = r \sin\theta$$ $$z = z$$ And the element of volume in cylindrical coordinates, \(dV\), is given by: $$dV = r dz dr d\theta$$ Now, let's substitute this into our integral: $$\int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}} \int_{0}^{9-3\sqrt{x^{2}+y^{2}}} d z d x d y = \int \int \int r dz dr d\theta$$

Step 2: Determine the new limits of integration

We need to find the new limits of integration in cylindrical coordinates that correspond to the original Cartesian coordinates limits. We can analyze the given limits as follows: Outer integral limit: \(0\leq y\leq 3\), it means we are dealing with a semi-circular cross-section in the \(xy\) plane. Since the semi-circle is symmetric about x-axis, \(0\leq \theta\leq \pi\). Middle integral limit: \(-\sqrt{9-y^2}\leq x\leq \sqrt{9-y^2}\), it represents a circle in the \(xy\) plane with radius 3, so we get \(0\leq r\leq 3\). Inner integral limit: \(0\leq z\leq 9-3\sqrt{x^2+y^2}\), no transformation required, since \(z=z\). Hence, our integral becomes: $$\int_{0}^{\pi} \int_{0}^{3} \int_{0}^{9-3\sqrt{r^2}} r dz dr d\theta$$

Step 3: Evaluate the integral

Now, let's evaluate the triple integral step-by-step. Inner integral: $$\int_{0}^{9-3\sqrt{r^2}}r dz = \left[\frac{1}{2}rz^2\right]_{0}^{9-3\sqrt{r^2}} = \frac{1}{2}r(9-3\sqrt{r^2})^2$$ Now our remaining double integral is: $$\int_{0}^{\pi} \int_{0}^{3} \frac{1}{2}r(9-3\sqrt{r^2})^2 dr d\theta$$ Middle integral: $$\int_{0}^{3} \frac{1}{2}r(9-3\sqrt{r^2})^2 dr = \left. \frac{1}{162}(18r^3+\sqrt{r^2}(729-81r^2)^{\frac{3}{2}}-729r^3) \right|_0^3$$ Which simplifies to: $$\frac{1}{162}(18\cdot 3^3+3\cdot 729 - 729\cdot 3^3) = \frac{729}{8}$$ Now our remaining single integral is: $$\int_{0}^{\pi} \frac{729}{8} d\theta$$ Outer integral: $$\int_{0}^{\pi} \frac{729}{8} d\theta = \frac{729}{8} \cdot \pi$$ So the value of the original triple integral is: $$\int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}} \int_{0}^{9-3\sqrt{x^{2}+y^{2}}} d z d x d y = \frac{729}{8} \cdot \pi$$

Key Concepts

Triple Integral

Triple integrals extend the concept of double integrals by considering functions of three variables and integrating over a three-dimensional region. They are useful for calculating the volume under a surface in a three-dimensional space and have applications in physics, engineering, and probability, among other fields. In our exercise, the triple integral is used to find the volume of a solid by integrating a constant function over the region bounded by the given limits of integration.

The process involves calculating the innermost integral first, followed by the middle integral, and finally the outermost integral, effectively collapsing the three-dimensional region into a single value representing the volume. It's important to remember that the order of integration (i.e., which variable to integrate first) can simplify or complicate the calculation, depending on the limits of integration and the geometry of the region.

Volume Element in Cylindrical Coordinates

In cylindrical coordinates, the volume element, denoted as \(dV\), changes from the Cartesian \(dx dy dz\) to \(r dr d\theta dz\). Cylindrical coordinates are used when the symmetry of the problem suggests that it would be easier to approach the problem using a cylindrical volume element rather than a rectilinear one.

In problems involving cylindrical symmetry, as in our given problem, using \(r dr d\theta dz\) simplifies the integral, particularly when the region of integration is a cylinder or part of a cylinder. The extra \(r\) factor in the volume element accounts for the circular cross sections increasing in circumference as one moves away from the origin, making it essential to include this term to calculate the correct volume.

Cylindrical Coordinate System

The cylindrical coordinate system is a three-dimensional coordinate system that extends the two-dimensional polar coordinate system with an extra dimension for height (z-coordinate). A point in cylindrical coordinates is represented by \((r, \theta, z)\), where:\

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• \(r\) is the radial distance from the origin,\
• \(\theta\) is the angle measured from the positive x-axis in the counter-clockwise direction,\
• \(z\) is the height above the xy-plane, similar to the z-coordinate in Cartesian coordinates.\

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This coordinate system is particularly well-suited for problems with rotational symmetry, such as those involving cylinders or cones. For example, in our exercise, the region of integration is a semi-cylindrical region, which is why converting to cylindrical coordinates is a natural and beneficial step.

Limits of Integration

The limits of integration in an integral define the region over which the integration is performed. In the context of cylindrical coordinates, it is often necessary to redefine the limits of integration from Cartesian to cylindrical to account for the changed geometry. This involves translating the bounds of x and y into bounds for \(r\) and \(\theta\), and usually, the bounds for z remain the same since it translates directly between Cartesian and cylindrical coordinates.

For instance, a circular limit in the xy-plane represented in Cartesian coordinates would have to be converted into radial bounds \(0\) to \(R\) and angular bounds \(0\) to \(2\theta\) in cylindrical coordinates. Our exercise's region of integration is a semi-cylinder, and so the limits are defined by a circle in the xy-plane and a linear function in z, which after translation into cylindrical coordinates becomes a region bounded by simple limits for \(r\), \(0\) to the radius of the cylinder, and for \(\theta\), \(0\) to \(\theta\), and for \(z\), from \(0\) to a linear function of \(r\).