#### To determine

**To:**

Sketch the region and find the enclosed area.

#### Answer

i) The sketch of the region

ii) Area =2215

#### 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 by xR and the left boundary by xL, then the area between these curves is given by

A=∫cdxR-xLdy

**2) Given:**

x=y4, y=2-x And y=0

**3) Calculation:**

Write both the equations as x in the terms of y.

y=2-x

y2=2-x

x=2-y2 --- (1)

x=y4 --- (2)

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

2-y2=y4.

y4+y2-2=0

y2+2y2-1=0

y=1 and y=-1

Exclude y=-1, because y≥0.

Thus, the points of intersection in the region under consideration is at y=1. The region is sketched in the following figure.

Here, the right curve is x=2-y2 and the left curve is x=y4. Therefore, let’s assume that

xR=2-y2

xL=y4

Therefore, the required area is

A=∫012-y2-y4dy

A=∫012-y2-y4 dy

Compute the integral using the standard integration rule.

A=2y-y33-y5501

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

A=21-133-155-0-0-0

A=2-13-15

A=2-815

A=2215

**Conclusion:**

i) The sketch of the region

ii) Area =2215