sketch the region enclosed by the given curves. decide whether to integrate with respect to x or y. then…

Explanation:

Step1: Solve equations for (x)

• For (2y = 4\sqrt{x}), we get (x=\frac{y^{2}}{4}).
• For (2y + 2x=6), we get (x = 3 - y).

Step2: Find intersection points

• Intersection of (y = 4) and (x=\frac{y^{2}}{4}): Substitute (y = 4) into (x=\frac{y^{2}}{4}), then (x=\frac{16}{4}=4).
• Intersection of (y = 4) and (x = 3 - y): Substitute (y = 4) into (x = 3 - y), then (x=-1).
• Intersection of (x=\frac{y^{2}}{4}) and (x = 3 - y): Set (\frac{y^{2}}{4}=3 - y), (y^{2}+4y - 12=0), ((y + 6)(y - 2)=0). So (y = 2) (since we consider the region with (y\geq0) from (2y = 4\sqrt{x})). When (y = 2), (x=\frac{4}{4}=1) from (x=\frac{y^{2}}{4}) and (x=3 - 2 = 1) from (x = 3 - y).

Step3: Set up the integral

Since it is easier to integrate with respect to (y) (the right - hand and left - hand functions of (x) are well - defined as functions of (y) over the interval of (y) values ([2,4])), the area (A=\int_{a}^{b}\left[\text{right - hand }x-\text{left - hand }x\right]dy). Here, the right - hand (x) is (\frac{y^{2}}{4}) and the left - hand (x) is (3 - y) for (y\in[2,4]). [ \begin{align*} A&=\int_{2}^{4}\left(\frac{y^{2}}{4}-(3 - y)\right)dy\ &=\int_{2}^{4}\left(\frac{y^{2}}{4}+y - 3\right)dy \end{align*} ]

Step4: Evaluate the integral

[ \begin{align*} \int_{2}^{4}\left(\frac{y^{2}}{4}+y - 3\right)dy&=\left[\frac{y^{3}}{12}+\frac{y^{2}}{2}-3y\right]_{2}^{4}\ &=\left(\frac{4^{3}}{12}+\frac{4^{2}}{2}-3\times4\right)-\left(\frac{2^{3}}{12}+\frac{2^{2}}{2}-3\times2\right)\ &=\left(\frac{64}{12}+8 - 12\right)-\left(\frac{8}{12}+2 - 6\right)\ &=\left(\frac{16}{3}-4\right)-\left(\frac{2}{3}-4\right)\ &=\frac{16 - 12}{3}-\frac{2 - 12}{3}\ &=\frac{4}{3}-\left(-\frac{10}{3}\right)\ &=\frac{4 + 10}{3}\ &=\frac{14}{3} \end{align*} ]

Answer:

(\frac{14}{3})