Question

Set up an integral for both orders of integration, and use the more convenient order to evaluate the integral over the region \(R\). \(\int_{R} \int-2 y d A\) \(R:\) region bounded by \(y=4-x^{2}, y=4-x\)

Short answer

The value of the double integral over the region R is -49/30.

Step-by-step solution

Find the Intersection Points Between the Curves

Set \(4 - x^{2} = 4 - x\) to find the points where the two curves intersect. Solving for x gives \(x = 0\) and \(x = 1\). This will serve as the x limits of integration.

Identify the Upper and Lower Functions

Between \(x = 0\) and \(x = 1\), \(4 - x^{2}\) is less than \(4 - x\), so the function \(4 - x^{2}\) is the lower function, and \(4 - x\) is the upper function over this interval.

Set the Integral and Determine the Correct Order of Integration

For this case, integrating first with respect to y gives the more convenient limits. So the integral is \(\int_{0}^{1}\int_{4-x^{2}}^{4-x}-2y \, dy \, dx\)

Evaluate the Inner Integral

The inner integral becomes \(-2 \int_{4-x^{2}}^{4-x}y \, dy\), which simplifies to \(-[y^{2}]_{4-x^{2}}^{4-x}\), and further simplifies to \(-(16 - 8x + x^{2} - 16 + 8x^{2} - x^{4})\).

Evaluate the Outer Integral

Now evaluate the outer integral: \(\int_{0}^{1}(x^{4} - 8x^{2} + x^{2}) \, dx\), which simplifies to \([1/5 x^{5} - 8/3 x^{3} + 1/3 x^{3}]_{0}^{1}\), and this simplifies further to -49/30. So the entire double integral over the region R becomes -49/30.