Question

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

Short answer

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 ≤ x ≤ 5 and -2 ≤ y ≤ 4, and the height of the box is 10 units. Answer: The volume of the box is 300 cubic units.

Step-by-step solution

Identify the limits of integration

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

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: $$V={\int }_0^{10}{\int }_{-2}^4{\int }_0^{5}1dxdydz$$

Solve the iterated integral

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