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Chapter 13: Problem 2

Write an iterated integral that gives the volume of a box with height 10 and base \(R=\\{(x, y): 0 \leq x \leq 5,-2 \leq y \leq 4\\}\)

Short Answer

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Question: Determine the volume of a box with a rectangular base and given height, where the base region in the xy-plane is defined by 0 鈮� x 鈮� 5 and -2 鈮� y 鈮� 4, and the height of the box is 10 units. Answer: The volume of the box is 300 cubic units.

Step by step solution

01

Identify the limits of integration

The coordinate limits provided for the rectangular base in the problem are: - 0 鈮� x 鈮� 5 - -2 鈮� y 鈮� 4 The height of the box is 10, which gives us the limits for z: - 0 鈮� z 鈮� 10 Now, with the limits for each coordinate, we can construct the triple integral.
02

Set up the iterated integral

A volume integral requires integrating with respect to all three coordinates (x, y, z), and the integral takes the following form: $$ V = \iiint_{R\times[0, 10]}dV $$ We can rewrite this integral as an iterated integral with respect to z, y, and x, in that order. So, the iterated integral for the volume V is: $$ V = \int_{0}^{10} \int_{-2}^{4} \int_{0}^{5} 1 ~dx~dy~dz $$
03

Solve the iterated integral

We will first integrate with respect to x, then y, and finally z. Integrate with respect to x: $$ \int_{0}^{5} 1~dx = x \Big|_0^5 = 5 $$ Next, integrate with respect to y, using the result of the previous integral: $$ 5\int_{-2}^{4} dy = 5(y \Big|_{-2}^4) = 5(4 - (-2)) = 30 $$ Finally, integrate with respect to z, using the previous result: $$ 30\int_{0}^{10} dz = 30(z \Big|_0^{10}) = 30(10-0) = 300 $$ Therefore, the volume of the box is 300 cubic units.

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