Find The Area Of The Region Bounded By The Curves, 𝑦=−6, 𝑦=6 The 𝑦−𝑎𝑥𝑖𝑠 And The Parabola 𝑦^2-4𝑥−16 = 0.

Question

Find The Area Of The Region Bounded By The Curves, 𝑦=−6, 𝑦=6 The 𝑦−𝑎𝑥𝑖𝑠 And The Parabola 𝑦^2-4𝑥−16 = 0.

My answer is -28/3. Just making sure

Answer 1

area = 9 1/3 units^2

You got it right 28/3 as it = 9 1/3

just drop the negative sign. a negative sign generally means an area below the x axis. In this problem the area is both above and below so you add them, using the absolute value of the negative area

y^2-4x-16 =0 is a rightward opening parabola

helps to sketch the graph. there are 2 separate areas,

one in quadrant I and a 2nd in quadrant IV

find area between y=-6, y=6 and the parabola y^2 -4x-16 =0

y^2 = 4x+16=4(x+4)

find one of the areas, then double it

area in quadrant I is 30 minus

area = integral of +/-2(x+4)^.5 evaluated from 0 to 5

= +2(x+4)^1.5/1.5 times 2 evaluated from 0 to 5

= (2/1.5)(x+4)^1.5 from x =0 to 5

= (4/3)(x+4)^(3/2) from 0 5o 5

= (4/3)(27 -8

= (4/3)(19)

=76/3 = 25 1/3

30 -25 1/3 = 4 2/3

double it to get

8 4/3

= 9 1/3

or

do the problem using the integral of an expression in terms of y

and get the same answer.

area = 2 times integral of (y^2/4 -4) dy

evaluated from 4 to 6

convert the 0 to 5 x limits y=2sqr(4x-16) = 2sqr(4(5)-16) =2sqr4 = 2*2 =4

find y value where parabola intersects y=6, that's y =6

integral of y^2/4 -4 = y^3/12 -4y

evaluate from 4 to 6

6^3/12 -4(6) - (4^3/12 -4(4)

= 18-24 - 16/3 +16

= 10 - 16/3 = 10-5 1/3

= 4 2/3

double it to get the total area below & above the x axis

= 8 4/3

= 9 1/3 square units

Answer 2

First, let's rewrite the equation as x in terms of y:

y² - 4x - 16 = 0

4x = y² - 16

x = y²/4 - 4

Now we can find the definite integral from -6 to 6:

∫_-6⁶ (y²/4 - 4)dy = (y³/12 - 4y) ]_-6⁶ = (6³/12 - 4*6) - ((-6)³/12 - 4*(-6)) = -12

So the area is -12.