Problem 22 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Evaluate the triple integral of the function f(x, y, z), where f(x, y, z) = (x^2 + y^2)^(1/2) over the region -1 ≤ z ≤ 1, 0 ≤ y ≤ 1/2, √(3y) ≤ x ≤ √(1-y^2) in cylindrical coordinates. Answer: The value of the integral is approximately 4/9.

Step 1: Convert function to cylindrical coordinates

First we need to convert the function f(x, y, z) = (x^2 + y^2)^(1/2) to cylindrical coordinates. We know that in cylindrical coordinates \((x, y, z) \to (\rho, \phi, z)\), where \(x = \rho \cos\phi\), \(y = \rho \sin\phi\), and \(z = z\). Then the function converts to \(f(\rho, \phi, z) = \rho\).

Step 2: Convert the limits of integration to cylindrical coordinates

Next, we need to rewrite the limits of integration in cylindrical coordinates. For \(z\): \(-1 \leq z \leq 1\) For \(y\): \(0 \leq y \leq \frac{1}{2} \implies 0 \leq \rho\sin\phi \leq \frac{1}{2}\) For \(x\): \(\sqrt{3y} \leq x \leq \sqrt{1-y^2} \implies \sqrt{3}\rho \sin\phi \leq \rho \cos\phi \leq \sqrt{1-\rho^2\sin^2\phi}\) Since \(\sqrt{1-\rho^2\sin^2\phi}\) is the maximum value for \(\rho\cos\phi\), we can solve for the upper limit of \(\rho\): $$\rho = \frac{1}{\sin\phi+\cos\phi}$$ Hence, the limits for \(\rho\) are: \(0 \leq \rho \leq \frac{1}{\sin\phi+\cos\phi}\) For \(\phi\), observe that the limits of \(x\) and \(y\) lead to: $$\phi = \arctan\left(\frac{\sqrt{3}}{1}\right) \implies 0 \leq \phi \leq \frac{\pi}{3}$$

Step 3: Plug the new limits of integration and function, and evaluate the integral

Now we rewrite the integral using the cylindrical coordinates: $$\int_{-1}^{1} \int_{0}^{\pi/3} \int_{0}^{\frac{1}{\sin\phi+\cos\phi}} \rho \cdot \rho\, d\rho\, d\phi\, dz = \int_{-1}^{1} \int_{0}^{\pi/3} \int_{0}^{\frac{1}{\sin\phi+\cos\phi}} \rho^2\, d\rho\, d\phi\, dz$$ First, integrate with respect to \(\rho\): $$\int_{0}^{\frac{1}{\sin\phi+\cos\phi}} \rho^2\, d\rho = \left[\frac{\rho^3}{3}\right]_{0}^{\frac{1}{\sin\phi+\cos\phi}} = \frac{1}{3(\sin\phi+\cos\phi)^3}$$ Now, integrate with respect to \(\phi\): $$\int_{0}^{\pi/3} \frac{1}{3(\sin\phi+\cos\phi)^3}\, d\phi = \frac{2}{9}$$ Finally, integrate with respect to \(z\): $$\int_{-1}^{1} \frac{2}{9}\, dz = \left[\frac{2z}{9}\right]_{-1}^{1} = \frac{4}{9}$$ Therefore, the value of the integral is approximately: $$\int_{-1}^{1} \int_{0}^{1 / 2} \int_{\sqrt{3} y}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right)^{1 / 2} d x d y d z = \boxed{\frac{4}{9}}$$