Question

Sketch the region \(R\) and evaluate the iterated integral \(\int_{R} \int f(x, y) d A .\) $$ \int_{0}^{6} \int_{y / 2}^{3}(x+y) d x d y $$

Short answer

After the last step, the final answer to the exercise is \(81\). This is the value of the iterated integral over the region \(R\).

Step-by-step solution

Sketch the region \(R\)

To visualize the region of integration, consider the limits of the inner integral, \(x\) varies from \(y/2\) to \(3\), and in the outer integral, \(y\) ranges from \(0\) to \(6\). Thus, the region \(R\) is a rectangular region in the xy-plane defined by \(0 \leq y \leq 6\) and \(y/2 \leq x \leq 3\).

Evaluate the inner integral

Let's evaluate the inner integral first by keeping \(y\) constant. Thus, integrate \(f(x,y) = x + y\) with respect to \(x\), from \(y/2\) to \(3\). This gives \(\int_{y/2}^{3}(x+y) dx = [(x^2)/2 + yx]_{y/2}^{3} = (9/2) - (y^2)/4 + 3y - y^2/2\).

Simplify the result from Step 2

Group like-terms obtained in the above expression to simplify the equation, giving \(9/2 - (3y^2)/4 + 3y\).

Evaluate the outer integral

Now, take the resulting function from step 3, and integrate with respect to \(y\), from \(0\) to \(6\). Thus, evaluate \(\int_{0}^{6} (9/2 - (3y^2)/4 + 3y) dy = [9y/2 - y^3/4 + (3y^2)/2]_{0}^{6}\), then substitute \(y = 6\) and \(y = 0\) and subtract to get the final answer.