Question

find the area of the region bounded by the parabola x=4-y^2 and y axis​

Answer 1

Answer

Given equations of curves are

x

2

=4y .......(1)

and, x=4y−2 ......(2)

Equation 1 represents a parabola which is open upward having vertex (0,0) and equation 2 represents a straight line.

On putting the value of 4y from equation 1 in equation 2, we get,

x=x

2

−2

x

2

−x−2=0

(x+1)(x−2)=0

x=−1,2

When x=−1, then from equation 1, y=

4

1

and when x=2, then from equation 1, y=1

Therefore, points of intersection of given curves are (−1,

4

1

) and (2,1).

Therefore,

Required area = Area of shaded region BOAB

=

−1

∫

2

[y(line)−y(parabola)]dx

=

−1

∫

2

[(

4

x+2

)−

4

x

2

]dx

=

4

1

−1

∫

2

(x+2−x

2

)dx

=

4

1

[(2+4−

3

8

)−(

2

1

−2+

3

1

)]

=

4

1

(6−

3

8

+2−

6

5

)

=

4

1

.

6

27

=

8

9

sq.units

Step-by-step explanation:

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Answer 2

We need to find area bounded by the parabola,

$\longrightarrow x=4-y^2$

and y axis, i.e., $x=0.$

Assume,

$\longrightarrow4-y^2\geq0$

$\longrightarrow y^2-4\leq0$

$\longrightarrow y^2\leq4$

$\longrightarrow y\in[-2,\ 2]$

Hence the area bounded will be given by,

$\displaystyle\longrightarrow A=\int\limits_{-2}^2\left(4-y^2\right)\ dy$

$\displaystyle\longrightarrow A=4\big[y\big]_{-2}^2-\dfrac{1}{3}\left[y^3\right]_{-2}^2$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{32}{3}}}$