#### To determine

**To sketch:**

The region enclosed by the given curve and find the area of the region.

#### Answer

i. The graph of the region enclosed by the given curves is

The graph with typical approximating rectangle is,

ii. The area of the region bounded by the given curve is A=643

#### Explanation

**1) Concept:**

i. The intersection of the two curves is obtained by solving the simultaneous equation of curves.

ii. Identify the boundaries of the region.

Formula-

The area of the typical rectangle is

xR-xL∆y where xR = theright boundary curve and xL= the left boundary curve.

The total area is

A= limn→∞∑i=1nxR-xL∆y = ∫abxR-xLdy

**2) Given:**

4x+y2=12 and x=y

3) **Calculation:**

The given curves can be written as

4x+y2=12 ⇒x=12-y24 =3-y24

and x=y

First, find the intersection of these two curves by solving the simultaneous equation:

that is,

3-y24=y

3-y24-y=0

Multiplying through out by -4,

y2+4y-12=0

Thus we get,

y=-6 or y=2

Therefore, the points of intersection of the two curves is -6, -6and (2, 2)

The graph of the region enclosed by the given curves is

Find the area by integrating with respect to y(since, we found boundary curves in terms of y and the restriction on y is calculated above, so it is easy to find the area by integrating with the respective y).

From the graph, the boundary curves are

xR=3-y24 and xL=y

The graph with the typical rectangle is

The area of the typical rectangle is

xR-xL∆y=(3-y24-y) ∆y

And the region lies between y=-6and y=2

So, the total area is

A= ∫abxR-xLdy

= ∫-11(3-y24-y)dy

Integrating,

= 3y-y312-y22-62

By using the upper and the lower limits,

=32-2312-222-3-6--6312--622

Simplifying,

6-812-2+18-18+18

Simplifying,

22-23

A=643

Therefore, the area of the region bounded by the given curves is A=643

**Conclusion:**

The graph of the region enclosed by the given curves is

The graph with the typical rectangle is

ii. The area of the region bounded by the given curves is A=643