Question

D is enclosed by curves \(y = {x^2}\& y = 3x\)

Short answer

Hence, answer is \(\frac{{243}}{8}\)

Step-by-step solution

Step 1:- The graph below shows the region enclosed by

\(y = {x^2}andy = 3x\)

The two curves intersect at point \((0,0)and(3,9)\)

Step 2:- Clearly from the figure, the lower end of the vertical line touches the upper end of vertical

\(y = 3x\)

\( \Rightarrow \)y varies from \({x^2} to 3x\)

Also, shaded region,\(0 \le x \le 3\)

Thus, the region D is described as:

\(D = \{ (x,y)/0 \le x \le 3,{x^2} \le y \le 3x\} \)

Step 3- Considering horizontal line in the shaded area as shown

\( \Rightarrow \)x varies from \(\frac{y}{3}to\sqrt y \)

Also, shaded region \(\begin{aligned}{l}0 \le y \le 9\\\end{aligned}\)

\(D = \left\{ {(x,y)\left| {0 \le y \le 9,\frac{y}{3} \le x \le \sqrt y } \right.} \right\}\)

Step 4:-Applying limit of integration

Hence, answer is \(\frac{{243}}{8}\)