Question

Write an iterated integral that gives the volume of the solid bounded by the surface \(f(x, y)=x y\) over the square \(R=\\{(x, y): 0 \leq x \leq 2,1 \leq y \leq 3\\}\)

Short answer

Answer: The volume of the solid is 8.

Step-by-step solution

Write down the given information

From the exercise, we are given the surface \(f(x, y) = xy\), and the region \(R = \{(x, y): 0 \leq x \leq 2, 1 \leq y \leq 3\}\).

Set up the iterated integral

Based on the given region \(R\), we will integrate with respect to \(x\) first, and then with respect to \(y\). The bounds for \(x\) are \(0 \leq x \leq 2\), and the bounds for \(y\) are \(1 \leq y \leq 3\). Therefore, our iterated integral will look like this: \[V = \int_{1}^3 \int_{0}^2 f(x, y) dxdy = \int_{1}^3 \int_{0}^2 xy\, dxdy\]

Integrate with respect to x

Let's now integrate with respect to \(x\): \[\int_{0}^2 xy\, dx = \frac{1}{2}x^2y\Big|_0^2 = (2^2y)\frac{1}{2} - (0^2y)\frac{1}{2} = 2y\]

Integrate with respect to y

Now we will integrate the result from step 3 with respect to \(y\): \[V = \int_{1}^3 2y\, dy = y^2\Big|_1^3 = (3^2) - (1^2) = 9 - 1 = 8\]

Write down the final answer

After integrating with respect to both \(x\) and \(y\), we found that the volume of the solid bounded by the surface \(f(x, y)=xy\) and the square \(R = \{(x, y): 0 \leq x \leq 2, 1 \leq y \leq 3\}\) is \(V = 8\).