`y=x^2`

`y=6-x`

The bounded region of the two equations is:

(a) To determine the area of the bounded region with respect to x, draw a vertical strip. (See attached image.)

In the figure, the top of the vertical strip touches the graph of `y=6-x` and its bottom touches the graph of `y = x^2`. Also, the bounded region starts at `x=-3` and ends at`x=2`. Applying the formula

`A = int_a^b (y_(_(upper)) - y_(_(lower)))dx`

the integral needed to compute the area of the bounded region is

`A=int_(-3)^2 (6-x - x^2) dx`

`A= (6x - x^2/2 - x^3/3)` `|_(-3)^2`

`A= (6*2 - 2^2/2 - 2^3/3) - (3*(-2)-(-3)^2/2-(-3)^3/3) `

`A=22/3 - (-27/2)`

`A=125/6`

**Therefore, the area of the bounded region is `125/6` square units.**

(b) To determine the area of the bounded region, draw a horizontal strip. (See attached image)

Notice that the right ends of the strips touch either the line or the curve. So to get the area, divide the bounded region into two. And, apply the formula:

`A = int_a^b (x_(right)- x_(l eft)) dy`

To express the integral in terms of y variable, isolate the x in each equation.

- `y = x^2`

`+-sqrty = x`

- `y = 6- x`

`6 - y = x`

For the lower region, both ends of the horizontal strip touch the graph of `x=+-sqrty` . Its bounded region starts at y=0 and ends at y = 4. And for the upper region, the right end of the strip touches the graph of `x=6-y` and its left end touches the graph of `x=-sqrty` . And its bounded region starts at y = 4 and ends at y = 9.

So the integral needed to compute the area of the bounded region is

`A = int_0^4(sqrt y - (-sqrty))dy + int_4^9 (6 - y - (-sqrty))dy`

`A = int_0^4 2sqrty dy + int_4^9 (6 -y + sqrty)dy`

`A=int_0^4 2y^(1/2)dy` `+` `int_4^9 (6-y+y^(1/2))dy`

`A = (4/3y^(3/2))` `|_0^4` `+` `(6y-1/2y^2+2/3y^(3/2))` `|_4^9`

`A = [4/3*4^(3/2) - 4/3*0^(3/2)] + [ (6*9 -1/2*9^2+2/3*9^(3/2))-(6-4-1/2*4^2 + 2/3*4^(3/2))]`

`A=(32/3-0)+ (63/2-64/3)`

`A = 125/6`

**Therefore, the area of the bounded region is `125/6` square units.**

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