Question

Find the area bounded by the parabola y^2=4-x and y^2 =4-4x

Answer 1

Answer:

Intersection point of the curves y

2

=4x and y=x

are (0,0) and (1,2)

Area of the the shaded region is given by:

A=∫

0

2

y−

4

y

2

⇒A=[

2

y

2

]

0

2

−[

12

y

3

]

0

2

⇒A=2−

3

2

=

3

4

sq. units

Answer 2

Given are the equations of two parabolas, Find the area bounded by them.

Explanation:

• We have the equations of parabolas as, $y^2=4-x,\ \ \ \ \ \ \ \ \ ->x_1=4-y^2 \\y^2=4-4x\ \ \ \ \ \ \ \ \ ->x_2=\frac{4-y^2}{4}$   -----(a)
• Then the points of intersection of the parabolas are, $4-x=4-4x\\->x=0\\->y^2=4\\->y=\pm 4\\->(0,4)\ (0,-4)$  
• Since the y-coordinate is changing at the points of intersection the area bounded is calculated by integrating 'x' with respect to 'y' for the changing coordinates as the limits.
• Hence from (a) we get the area as,
• $A=\int\limits^4_{-4} {x_1-x_2} \, dx\ \\A=\int\limits^4_{-4} { 4-y^2 - \frac{4-y^2}{4}} \, dy \\A=\frac{3}{4} \int\limits^4_{-4} {(4-y^2)} \, dy \\A=\frac{3}{4} [4y-\frac{y^3}{3} ]_{-4}^4 \\\\A=|\frac{3}{4}[-\frac{32}{3} ] |\\->A=8\ sq.units$      
• The area bounded by the two parabolas is $8$ sq. units