Problem 19 - Calculus Early Transcendentals

Question

Evaluate the following integrals in cylindrical coordinates. $$\int_{0}^{4} \int_{0}^{\sqrt{2} / 2} \int_{x}^{\sqrt{1-x^{2}}} e^{-x^{2}-y^{2}} d y d x d z$$

Step-by-Step Solution

Answer

The given triple integral in cylindrical coordinates is equal to $$\frac{1}{2}\left(1-e^{-\frac{1}{\cos^2{\theta}}}\right) \left(\pi - 2 J\right)$$ where \(J = \int_{0}^{\frac{\pi}{4}} e^{-\frac{1}{\cos^2{\theta}}} d\theta\) is a non-elementary integral. The exact value of J is not known, so the value of the integral cannot be expressed in a closed form. However, if needed, J can be approximated using a numerical method like Simpson's rule, and the approximate value of the triple integral can be calculated.

Step 1: Convert the integrand to cylindrical coordinates

We first convert the given integrand from Cartesian to cylindrical coordinates. In cylindrical coordinates, \((x, y, z) \to (r, \theta, z)\), where \(x = r\cos{\theta}\), \(y = r\sin{\theta}\), and \(z = z\). Therefore, the exponential function in the integral becomes: $$e^{-x^{2}-y^{2}} = e^{-r^{2}\cos^2{\theta}-r^{2}\sin^2{\theta}} = e^{-r^{2}(\cos^2{\theta}+\sin^2{\theta})} = e^{-r^2}$$ In addition, we must include the Jacobian when transforming coordinates, which is given by \(r\) in cylindrical coordinates.

Step 2: Convert the limits of integration

Next, convert the limits of integration from Cartesian to cylindrical coordinates. For the current limits of integration: - \(0 \leq z \leq 4\) remains the same for z. - \(0 \leq x \leq \sqrt{2} / 2\) implies \(\sqrt{2} / 2 \geq r\cos{\theta}\); combined with the inequality \(r \geq 0\), we have that \(0 \leq r \leq \sqrt{2} / 2 / \cos{\theta}\). - \(x \leq y \leq \sqrt{1-x^{2}}\) implies that \(r\cos{\theta} \leq r\sin{\theta} \leq \sqrt{1-r^{2}\cos^2{\theta}}\). Therefore, the limits of integration become: $$0 \leq z \leq 4,\ 0 \leq r \leq \frac{\sqrt{2}}{2\cos{\theta}},\ r\cos{\theta} \leq r\sin{\theta} \leq \sqrt{1-r^{2}\cos^2{\theta}}$$

Step 3: Evaluate the triple integral in the new coordinate system

Now, we can rewrite the given integral as: $$\int_{0}^{4} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{\sqrt{2}}{2\cos{\theta}}} re^{-r^2} d r d\theta d z$$ Since the exponential does not depend on the angle \(\theta\), we can evaluate each of the integrals separately.

Step 4: Evaluate the r-integral

Use substitution to compute the integral with respect to r: $$I_r = \int_{0}^{\frac{\sqrt{2}}{2\cos{\theta}}} re^{-r^2} dr$$ Let \(u=r^2\), so \(du = 2r dr\). Then, $$I_r = \frac{1}{2}\int_{0}^{\frac{1}{\cos^2{\theta}}} e^{-u} du$$ Using the integral form of the exponential function, the integral becomes: $$I_r = \frac{1}{2}\Big[-e^{-u}\Big]_{0}^{\frac{1}{\cos^2{\theta}}} = \frac{1}{2}\Big(1-e^{-\frac{1}{\cos^2{\theta}}}\Big)$$

Step 5: Evaluate the \(\theta\)-integral

Now we need to compute the \(\theta\)-integral: $$I_{\theta} = \int_{0}^{\frac{\pi}{4}} \frac{1}{2}\Big(1-e^{-\frac{1}{\cos^2{\theta}}}\Big) d\theta$$ Split into two separate integrals and evaluate them: $$I_{\theta} = \frac{1}{2} \left[\theta\right]_{0}^{\frac{\pi}{4}} - \frac{1}{2} \int_{0}^{\frac{\pi}{4}} e^{-\frac{1}{\cos^2{\theta}}} d\theta$$ $$I_{\theta} = \frac{\pi}{8} - \frac{1}{2} \int_{0}^{\frac{\pi}{4}} e^{-\frac{1}{\cos^2{\theta}}} d\theta$$ The last integral does not have an elementary antiderivative, so denote it as J: $$J = \int_{0}^{\frac{\pi}{4}} e^{-\frac{1}{\cos^2{\theta}}} d\theta$$

Step 6: Evaluate the z-integral

Finally, we need to compute the z-integral and assemble the solution for the triple integral: $$I_z = \int_0^4 I_r I_{\theta} dz = \int_0^4 \frac{1}{2}\Big(1-e^{-\frac{1}{\cos^2{\theta}}}\Big) \left(\frac{\pi}{8} - \frac{1}{2} J\right) dz$$ $$I_z = \frac{1}{2}\left[\Big(1-e^{-\frac{1}{\cos^2{\theta}}}\Big) \left(\frac{\pi}{8}z - \frac{1}{2} Jz\right)\right]_0^4$$ $$I_z = \frac{1}{2}\left(1-e^{-\frac{1}{\cos^2{\theta}}}\right) \left(\pi - 2 J\right)$$ Thus, the given triple integral in cylindrical coordinates is: $$\boxed{\int_{0}^{4} \int_{0}^{\sqrt{2} / 2} \int_{x}^{\sqrt{1-x^{2}}} e^{-x^{2}-y^{2}} d y d x d z = \frac{1}{2}\left(1-e^{-\frac{1}{\cos^2{\theta}}}\right) \left(\pi - 2 J\right)}$$ where \(J = \int_{0}^{\frac{\pi}{4}} e^{-\frac{1}{\cos^2{\theta}}} d\theta\) is a non-elementary integral that can be approximated using a numerical method if required.

Key Concepts

Triple Integrals

Triple integrals are an extension of double integrals into three dimensions, allowing us to calculate the volume or mass of a 3D region. Consider each small volume element in a space being summed up. This method is useful in physics and engineering for computing various properties that depend on volume.
In the provided problem, converting to cylindrical coordinates simplifies the evaluation. Here, we integrate a function, involving the variables \(x\), \(y\), and \(z\), over a defined region. This is the essence of a triple integral where we go step by step across each coordinate to achieve the total integral.

Integration Techniques

Integration techniques refer to methods used to solve integrals efficiently. For non-standard regions or functions, transforming the integral into a different coordinate system, such as polar, cylindrical, or spherical, can simplify solving it.
In the exercise, the transformation from Cartesian to cylindrical coordinates simplifies the problem by leveraging the symmetry of the exponential function in \(e^{-x^2-y^2}\). The integration over \(r\), \(\theta\), and \(z\) coordinates aligns with the cylindrical nature of the region and the integrand, allowing for easier computation.

• Changing variables needs the Jacobian, here \(r\), to account for the transformation's scale.
• Substitution and parts are common core techniques for handling the integral after transformation.

Coordinate Transformation

Coordinate transformations convert integrals from one coordinate system to another, often simplifying limits or the integrand. In cylindrical coordinates, positions are described with the radius \(r\), angle \(\theta\), and height \(z\), which are direct enhancements over Cartesian coordinates for certain symmetrical or spiral-shaped regions.
In this case, \(x = r\cos\theta\) and \(y = r\sin\theta\), reduce the original problem's complexity by collapsing terms and specific limits. The exponential function simplifies to \(e^{-r^2}\) because the trigonometric terms sum to 1.

Limits of Integration

Setting accurate limits of integration is crucial in evaluating triple integrals. These set the bounds for each variable and describe the region over which the function is integrated. Different coordinate systems simplify this process by making limits more geometric or intuitive.
For cylindrical coordinates in this exercise:

• The \(z\)-bounds remain \(0 \leq z \leq 4\) as height doesn't depend on other variables.
• The radius \(r\) is bounded by the conversion from \(0 \leq x \leq \frac{\sqrt{2}}{2}\) and the angular relationship, \(0 \leq r \leq \frac{\sqrt{2}}{2\cos\theta}\).
• The angle \(\theta\) ranges from \(0\) to \(\frac{\pi}{4}\), determined by the region symmetry.

Non-elementary Integrals

Non-elementary integrals occur when the antiderivative of a function cannot be expressed using elementary functions like polynomials, exponentials, or simple trigonometric functions. They often require numerical approximation or special functions for solutions.
In this problem, \( J = \int_{0}^{\frac{\pi}{4}} e^{-\frac{1}{\cos^2{\theta}}} d\theta \) does not have an elementary antiderivative. It's computed or approximated using techniques such as numerical integration, like Simpson's rule or software tools.