Let's break down and solve the given problem **step by step**: --- ### **Problem Statement** Consider the iterated integral in cylindrical coordinates: \[ \int_{0}^{\pi/5} \int_{0}^{1} \int_{r}^{1} r \, dz \, dr \, d\theta \] You are to **plot the solid region** defined by these limits using MATLAB's `trisurf` function. --- ## **Step 1: Interpret the Integral Limits** In cylindrical coordinates: - \( r \) = radius (distance from \( z \)-axis) - \( \theta \) = angle - \( z \) = height The limits are: - \( 0 \leq \theta \leq \frac{\pi}{5} \) - \( 0 \leq r \leq 1 \) - \( r \leq z \leq 1 \) **This region is:** - A "wedge" of a cylinder from \(\theta = 0\) to \(\theta = \frac{\pi}{5}\) - With radius from 0 to 1 - For each \((r, \theta)\), \(z\) runs from \(r\) up to 1 (i.e., above the cone \(z = r\) and below the plane \(z = 1\)) --- ## **Step 2: Parameterize the Region** Convert cylindrical to Cartesian: \[ \begin{align*} x &= r\cos\theta \\ y &= r\sin\theta \\ z &= z \end{align*} \] So for all \((r, \theta)\) in the region: - \( x = r\cos\theta \) - \( y = r\sin\theta \) - \( z \) from \( r \) to \( 1 \) --- ## **Step 3: MATLAB Plotting Approach** ### **A. Surface 1: Bottom Surface \(z = r\)** - For \( r \) from 0 to 1, \( \theta \) from 0 to \(\pi/5\), \( z = r \) ### **B. Surface 2: Top Surface \(z = 1\)** - For \( r \) from 0 to 1, \( \theta \) from 0 to \(\pi/5\), \( z = 1 \) ### **C. Side Surfaces** - At \( r = 1 \): \( z \) goes from 1 to 1 (degenerate edge) - At \( \theta = 0 \) and \( \theta = \pi/5 \): vertical faces --- ## **Step 4: MATLAB Script Example** Here's a MATLAB script to plot the solid region using `surf` (for smooth surfaces) and `trisurf` if you want to triangulate the faces: ```matlab % Define grid for r and theta r = linspace(0,1,40); theta = linspace(0,pi/5,40); [R, Theta] = meshgrid(r, theta); % Bottom surface: z = r X1 = R .* cos(Theta); Y1 = R .* sin(Theta); Z1 = R; % Top surface: z = 1 X2 = X1; Y2 = Y1; Z2 = ones(size(R)); % Plot surfaces figure; hold on % Plot bottom surface (z = r) surf(X1, Y1, Z1, 'FaceAlpha', 0.5, 'EdgeColor','none'); % Plot top surface (z = 1) surf(X2, Y2, Z2, 'FaceAlpha', 0.5, 'EdgeColor','none'); % Plot side at theta = 0 X3 = r .* cos(0); Y3 = r .* sin(0); Z3 = [r; ones(size(r))]; fill3([X3, X3], [Y3, Y3], Z3, 'g', 'FaceAlpha', 0.5); % Plot side at theta = pi/5 X4 = r .* cos(pi/5); Y4 = r .* sin(pi/5); Z4 = [r; ones(size(r))]; fill3([X4, X4], [Y4, Y4], Z4, 'r', 'FaceAlpha', 0.5); xlabel('x'); ylabel('y'); zlabel('z'); title('Solid Region for Given Cylindrical Integral'); axis equal view(3) hold off ``` --- ## **Step 5: Final Explanation** - The solid described is the region **above the cone \(z=r\)**, **below the plane \(z=1\)**, **inside the cylinder \(r=1\)**, and **between the planes \(\theta=0\) and \(\theta=\pi/5\)**. - The plot in MATLAB will show a "wedge" from \(z=r\) up to \(z=1\) in the specified angular region. --- ## **Final Answer** **The region is the set of all \((x,y,z)\) such that:** \[ 0 \leq \theta \leq \frac{\pi}{5},\quad 0 \leq r \leq 1, \quad r \leq z \leq 1 \] where \(x = r\cos\theta\), \(y = r\sin\theta\). Use the MATLAB code above to visualize the region with appropriate axis labels. --- **If you want the code for triangulating and using `trisurf`, let me know!**

Let's carefully evaluate the volume \(V\) of the solid region described by the integral: \[ V = \int_{0}^{\pi/5} \int_{0}^{1} \int_{r}^{1} r \, dz \, dr \, d\theta \] --- ### **Step 1: Understand the integral** - The order of integration: \( dz \), then \( dr \), then \( d\theta \). - The limits: - \( \theta \in [0, \pi/5] \) - \( r \in [0, 1] \) - \( z \in [r, 1] \) - The integrand: \( r \) --- ### **Step 2: Integrate with respect to \(z\)** \[ \int_{z=r}^{1} r \, dz = r \times (1 - r) \] because \(r\) is treated as a constant during the \(z\)-integration. --- ### **Step 3: Now the integral reduces to** \[ V = \int_{0}^{\pi/5} \int_{0}^{1} r (1 - r) \, dr \, d\theta \] --- ### **Step 4: Integrate with respect to \(r\)** \[ \int_{0}^{1} r (1 - r) \, dr \] Let's compute this integral: \[ \int_{0}^{1} (r - r^2) \, dr = \left[ \frac{r^2}{2} - \frac{r^3}{3} \right]_0^1 = \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \] --- ### **Step 5: Integrate with respect to \(\theta\)** \[ V = \int_{0}^{\pi/5} \frac{1}{6} \, d\theta = \frac{1}{6} \times \left[ \theta \right]_0^{\pi/5} = \frac{1}{6} \times \frac{\pi}{5} = \frac{\pi}{30} \] --- ### **Final answer:** \[ \boxed{ \boxed{ V = \frac{\pi}{30} } } \] **or equivalently,** \[ V = \frac{\pi}{30} \] which numerically approximates to: \[ V \approx 0.10472 \] --- ### **Summary:** - The volume of the solid region is \(\boxed{\frac{\pi}{30}}\). - The integral evaluates neatly to this simple form after straightforward integration. **Therefore, the volume \(V = \boxed{\frac{\pi}{30}}\).**