Problem 4 - Calculus Early Transcendentals

Question

Sketch the region of integration for the integral \(\int_{0}^{1} \int_{0}^{\sqrt{1-z^{2}}} \int_{0}^{\sqrt{1-y^{2}-z^{2}}} f(x, y, z) d x d y d z\)

Step-by-Step Solution

Answer

Question: Sketch the region of integration for the given triple integral, with the limits of integration as follows: - For the variable z: \(0 \le z \le 1\). - For the variable y: \(0 \le y \le \sqrt{1-z^2}\). - For the variable x: \(0 \le x \le \sqrt{1-y^2-z^2}\). Answer: The region of integration is the volume enclosed by a quarter of a sphere with a radius of 1, centered at the origin in the first octant, a semicircle in the xy-plane, and a horizontal plane at z=1.

Step 1: Identify Limits of Integration

First, we identify the limits of integration for each variable: - For the variable z, the limits are \(0 \le z \le 1\). - For the variable y, the limits are \(0 \le y \le \sqrt{1-z^2}\). - For the variable x, the limits are \(0 \le x \le \sqrt{1-y^2-z^2}\).

Step 2: Analyze the z-boundaries

We analyze the z-boundaries. The variable z is given from 0 to 1 (the entire range in the z-axis). So, we have a range going from the xy-plane (for z=0) to the z=1 plane.

Step 3: Analyze the y-boundaries

Next, we analyze the y-boundaries. The variable y is defined in the range \(0 \le y \le \sqrt{1-z^2}\). When z = 0, we have \(0 \le y \le \sqrt{1}\), meaning that the variables y lie in the range from 0 to 1 (all along the xz-plane). As the value of z increases, the range of the y-coordinate will decrease and restrict more. For z=1, the y-range will be only at the origin y=0. Consequently, this range of y forms a semicircle in the xy-plane.

Step 4: Analyze the x-boundaries

Lastly, we analyze the x-boundaries. The variable x is given in the range of \(0 \le x \le \sqrt{1-y^2-z^2}\). In a similar way as we analyzed the y-boundaries, for the x-coordinate, when z=0 and y=0, the range is \(0 \le x \le \sqrt{1}\), meaning that the x variable lies in the range from 0 to 1 (on the yz-plane). As the value of y and z increase, the x-range will become more restricted. When y^2+z^2 = 1, then the x-boundary will be only at the origin x=0, forming a quarter of a sphere centered at the origin.

Step 5: Draw the region

With the information gathered from analyzing the boundaries, we can now sketch the region of integration. 1. Start with a 3D Cartesian coordinate system. 2. For the z-boundaries, draw the xy-plane and a horizontal plane at z=1. 3. For the y-boundaries, draw a semicircle in the xy-plane with a radius of 1, centered at the origin. 4. For the x-boundaries, draw a quarter of a sphere centered at the origin with a radius of 1, inside the first octant (where x, y, and z are all positive). 5. The region of integration is the volume enclosed by the quarter of the sphere, the semicircle in the xy-plane, and the z=1 plane. With this sketch, we've successfully visualized the region of integration for the given triple integral.