# Compute the area between `x+y=16 and x+4= y^2` . Show the 2 ways of finding the area bounded by the two curves.

lemjay | High School Teacher | (Level 3) Senior Educator

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Let,

EQ1: `x+y=16`               and              EQ2: `x+4=y^2`

Then, solve the intersection points of EQ1 and EQ2. Use elimination method of system of equations.

So, subtract EQ2 from EQ1.

`x+y=16`

(-)   `x+4=y^2`

`----------`

`0+y-4=16-y^2`

`y-4=16-y^2`

`y^2+y-4-16=0`

`y^2+y-20=0`

`(y+5)(y-4)=0`

Set each factor equal to zero.

`y+5=0`                and            `y-4=0`

`y=-5`                                    `y=4`

Substitute values of y to either EQ1 or EQ2.

`y=-5 ` ,         `x-5=16`              `x=21`

`y=4` ,            `x+4=16`              `x=12`

So, EQ1 and EQ2 intersects at (12,4) and (21,-5).

And the area bounded by the two curves are:

(Note: Blue-Graph of EQ1 and Black-Graph of EQ2.)

To solve for the area between two curves, either of these formulas can be used.

(I) `A= int_a^b (y_U-y_L)dx`

(II) `A=int_a^b (x_R-x_L)dy`

(I) If we use the first formula, based on the graph, there are two upper functions which are the positive portion of the parabola                            (`y_(U1)=sqrt(x+4)` ) and the line (`Y_(U2)=16-x` ). And there is one lower function which is the negative portion of the parabola (`y_L=-sqrt(x+4)`).

So we have:

`A= int_a^b (sqrt(x+4)-(-sqrt(x+4)))dx + int_b^c (16-x - (-sqrt(x+4))dx)`

`A=2int_a^b sqrt(x+4)dx + int_b^c (16-x+sqrt(x+4)dx`

Note that the limits of the integral a, b,c are the x-coordinates of the vertex(-4,0), and the two intersection points (12,4) and (21,-5) respectively.

`A=2int_(-4)^12 (x+4)^(1/2)dx + int_12^21 (16-x+(x+4)^(1/2)dx`

`A=4/3(x+4)^(3/2)`   `|_(-4)^21`   `+`  `(16x-x/2+2/3(x+4)^(1/2))`  `|_12^21`

`A = 256/3 + 217/6`

`A = 243/2 `

Hence, the area between the given curves is `243/2` square units.

(II) If we used the second formula, based on the graph the function at the right is the line (`x_R=16-y` ) and the function at the left is the parabola (`x_L=y^2-4` ).

So we have,

`A=int_alpha^beta (16-y-(y^2-4))dy = int_alpha^beta (20-y-y^2)dy`

Here, the limits of the integral are the y-coordinates of the intersection points (21,-5) and (12,4).

`A= int_(-5)^4 (20-y-y^2)dy = 20y-y^2/y-y^3/3`  `|_(-5)^4`

`A = 243/2`

Hence, the area between the given curves is `243/2` square units.

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