#### To determine

**To:**

Sketch the region and find the enclosed area

#### Answer

i) The sketch of the region

ii) Area =323

#### Explanation

**1) Concept:**

Formula:

The area A of the region bounded by the curves x=f(y), x=g(y) and the lines y=c and y=d is

A= ∫cdfy-gydy

Let’s denote the right boundary curve by xR and the left boundary by xL, then the area between these curves is given by

A=∫cdxR-xLdy

**2) Given:**

x=2y2 and x=4+y2

**3) Calculation:**

The point of intersection occurs when both equations are equal to each other, that is,

2y2=4+y2

y2=4

Thus, the points of intersection are at y=-2 and y=2. The region is sketched in the following figure.

Here, the right curve is x=4+y2 and the left curve is x=2y2. Therefore,

xR=4+y2.

xL=2y2

Therefore, the required area is

A=∫-224+y2-2y2dy

A=∫-224+y2-2y2 dy

After simplifying this, it becomes

A=∫-22(4-y2)dy

Using the sum and difference rule and the constant multiple rule of integral, it becomes

A=4∫-22dy-∫-22y2 dy

Compute the integral using the standard integration rule.

A=4y-22-y33-22

Evaluate the integral by plugging the upper and the lower limits of integration.

A=42--2-233--233

A=42+2-83+83

A=4·4-163

A=48-163

A=323

**Conclusion:**

i) The sketch of the region

ii) Area =323