# Find the number b such that the line y = b divides the region bounded by the curves y = (x^2)and y = 4 into two regions of equal area. Set up and evaluate the integrals to do this by both...

Find the number b such that the line y = b divides the region bounded by the curves y = (x^2)and y = 4 into two regions of equal area.

Set up and evaluate the integrals to do this by both horizontal and vertical slicing. Sketch the regions.

### 2 Answers | Add Yours

Consider the graphs of the question:

To properly evaluate this, we need to find the line y=b so that the divided regions are of equal area. This means that we need to integrate using dy and horizontal slicing instead of integrate using dx.

The function `y=x^2` now becomes the two functions `x=+-sqrt y` .

A slice of width dy becomes:

`dA=(sqrt y - (-sqrt y) )dy`

`=2sqrt y dy`

and this is integrated as

`A=int_0^b2sqrt y dy+int_b^4 2sqrt y dy` evaluate each integral separately:

`=2 (2/3)(y^(3/2))_0^b+2 (2/3)(y^(3/2))_b^4` but these two parts must be equal, so we can solve for b.

`4/3 b^(3/2)=4/3(4^{3/2}-b^(3/2))` divide by 4/3 and simplify

`2b^(3/2)=4^(3/2)` evaluate right side

`b^(3/2)=4` square both sides

`b^3=16` take cube roots

`b=16^{1/3} approx 2.51`

**Using horizontal slicing, we find the line that divides the region is **`y=16^{1/3}`

To evaluate this using vertical slicing we consider the vertical slices for the top region as:

`dA=(4-b)dx` from `-sqrt b` to `sqrt b`

and

`dA=(4-x^2)dx` from `-2` to `-sqrt b`

and

`dA=(4-x^2)dx` from `sqrt b` to `2`

For the bottom region, we have

`dA=b dx` from `-sqrt b` to `sqrt b`

Which gives the integrals:

`int_{-sqrt b}^{sqrt b} b dx = int_{-2}^{-sqrt b}(4-x^2)dx+int_{-sqrt b}^{sqrt b}(4-b)dx+int_{sqrt b}^2(4-x^2)dx`

`bx_{-sqrt b}^{sqrt b}=2int_{sqrt b}^2(4-x^2)dx+2int_0^{sqrt b}(4-b)dx`

`2b^{3/2}=2(4x-1/3x^3)_{sqrtb}^2+2(4 sqrtb-b^{3/2})` divide by 2 continue to simplify

` ` `b^{3/2}=8-4sqrtb-8/3+b^{3/2}/3+4sqrt b-b^{3/2}` multiply by 3, continue to simplify

` ` `b^{3/2}=4` which leads to the same result as earlier.

**Using horizontal or vertical slicing, we find the line that divides the region is **`y=16^{1/3}`

`y=4` and `y=x^2` intersect when `x^2=4` , so `x= +-2`

Thus the area of the region is: `int_(-2)^2 (4-x^2) dx`

This is a symmetric region, so we can just do: `2 int_0^2 (4-x^2)dx` instead

`=2[ 4x-(x^3)/3 |_0^2 ]`

`=2[(8-8/3)-(0-0)] =32/3 `

So we want to find the horizonal line that splits the region into two regions with area 16/3

If we had such a b, then we would be looking where `y=b` intersects `y=x^2`

This happens at `x=+- sqrt(b)`

So we want to find `b` such that `int_(-sqrt(b))^(sqrt(b)) (b-x^2)dx = 16/3`

As before, we can use symmetry:

`2 int_0^(sqrt(b)) (b-x^2) dx = 16/3`

`int_0^(sqrt(b)) (b-x^2) dx=8/3`

`[bx-(x^3)/3 |_0^(sqrt(b)) ] = 8/3`

`(bsqrt(b)-(bsqrt(b))/3)-(0-0) = 8/3`

`(2/3) b sqrt(b) = 8/3`

`bsqrt(b) = 4`

`b^(3/2) = 4`

`b=4^(2/3)`

So the horizontal line that splits the region into two equal areas is:`` `y=4^(2/3)`

That was with vertical slicing.

To do this with horizontal slicing, again we observe that the region is symmetric, so we can just concentrate on the right half, and then multiply by 2. So we focus on:

A horizontal slice is now given by `(sqrt(y) - 0)` , the thickness of a slice is `dy` and the slices go from 0 to 4. Thus we have:

`int_0^4 sqrt(y) dy = int_0^4 y^(1/2)dy =[ (2/3)y^(3/2)] |_0^4 = (2/3) [4^(3/2) - 0] = 16/3`

(The actual region is double this, so the actual area is 32/3)

Now we want to find the horizontal line that cuts the region in half. Thus we want:

`int_0^b sqrt(y) dy = 8/3`

So:

`int_0^b y^(1/2) dy = [(2/3)y^(3/2)] |_0^b = (2/3) [b^(3/2) - 0] = 8/3`

`2/3 b^(3/2) = 8/3`

`b^(3/2) = (8/3)(3/2) = 4`

`b = 4^(2/3) ~~2.5198`