Problem 21 - Calculus Early Transcendentals

Question

Evaluate the following integrals in cylindrical coordinates. $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \int_{0}^{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}\right)^{-1 / 2} d z d y d x$$

Step-by-Step Solution

Answer

Question: Evaluate the triple integral in cylindrical coordinates: $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \int_{0}^{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}\right)^{-1 / 2} d z d y d x$$ Answer: The value of the given triple integral in cylindrical coordinates is \(\frac{9\pi}{4}\).

Step 1: Convert to Cylindrical Coordinates

To convert the given integral into cylindrical coordinates, we'll replace the Cartesian coordinates (x, y, z) with cylindrical coordinates (ρ, φ, z) using the following transformations: - \(x = ρ\cos{φ}\) - \(y = ρ\sin{φ}\) - \(z = z\) Additionally, the Jacobian for the transformation between Cartesian and cylindrical coordinates is \(ρ\). So, the integral becomes: $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \int_{0}^{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}\right)^{-1 / 2} \rho d z d y d x$$ Now, let's plug in the cylindrical coordinates conversions and change the integral limits.

Step 2: Change Limits of Integration

After converting to cylindrical coordinates, we rewrite the limits of integration: - For \(x\), we know that \(x = \rho\cos{φ}\), and the integral limits are from \(0\) to \(3\). So, \(\rho\cos{φ}\) ranges from \(0\) to \(3\). Therefore, we have \(ρ\) ranges from \(0\) to \(3\). - For \(y\), we know that \(y = \rho\sin{φ}\), and the integral limits are from \(0\) to \(\sqrt{9-x^{2}}\). So, \(\rho\sin{φ}\) ranges from \(0\) to \(\sqrt{9-(\rho\cos{φ})^{2}}\). This implies that \(φ\) ranges from \(0\) to \(\pi/2\). - For \(z\), the integral limits are from \(0\) to \(\sqrt{x^{2}+y^{2}} = \rho\). With the new limits and the Jacobian, the integral becomes: $$\int_{0}^{3} \int_{0}^{\pi/2} \int_{0}^{\rho}\rho^{-1} \rho d z d\phi d\rho$$

Step 3: Simplify the Integrals

Simplify the integral: $$\int_{0}^{3} \int_{0}^{\pi/2} \int_{0}^{\rho} d z d\phi d\rho$$ Now, we can solve the integral for \(z\), \(φ\), and \(ρ\) one by one.

Step 4: Integrate with respect to z

First, solve the inner integral with respect to \(z\): $$\int_{0}^{3} \int_{0}^{\pi/2} \left[z\right]_{0}^{\rho} d\phi d\rho = \int_{0}^{3} \int_{0}^{\pi/2} \rho d\phi d\rho$$

Step 5: Integrate with respect to φ

Next, solve the middle integral with respect to \(φ\): $$\int_{0}^{3} \left[\rho\phi\right]_{0}^{\pi/2} d\rho = \int_{0}^{3} \frac{\pi}{2}\rho d\rho$$

Step 6: Integrate with respect to ρ

Finally, solve the outer integral with respect to \(ρ\): $$\left[\frac{\pi}{4}\rho^2\right]_{0}^{3} = \frac{9\pi}{4} - 0 = \boxed{\frac{9\pi}{4}}$$ The value of the given integral in cylindrical coordinates is \(\frac{9\pi}{4}\).