# Find b s.t. y=b divides area between y=25-x^2, y=0 into 2 equal halves

#### Seed5813

##### New member
1. LarCalcET6 7.1.069.

Find b such that the line y = b divides the region bounded by the graphs of y = 25 - x[SUP]2[/SUP] and y = 0 into two regions of equal area.

[HR][/HR]
I got:

. . . . .$$\displaystyle \displaystyle \int_0^5\, (25\, -\, x^2)\, dx\, =\, \dfrac{250}{3}$$

Then tried to find:

. . . . .$$\displaystyle \displaystyle \int_0^b\, (25\, -\, x^2)\, dx\, =\, \dfrac{125}{3}$$

But I get stuck when I try to evaluate it:

. . . . .$$\displaystyle \displaystyle 3\int_0^b\, (25\, -\, x^2)\, =\, 125$$

. . . . .$$\displaystyle \displaystyle 3\left[\, 25x\, -\, \frac{1}{3}x^3\right]_0^b\, =\, 125$$

. . . . .$$\displaystyle \displaystyle 75b\, -\, b^3\, =\, 125$$

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#### Dr.Peterson

##### Elite Member
1. LarCalcET6 7.1.069.

Find b such that the line y = b divides the region bounded by the graphs of y = 25 - x[SUP]2[/SUP] and y = 0 into two regions of equal area.

[HR][/HR]
I got:

. . . . .$$\displaystyle \displaystyle \int_0^5\, (25\, -\, x^2)\, dx\, =\, \dfrac{250}{3}$$

Then tried to find:

. . . . .$$\displaystyle \displaystyle \int_0^b\, (25\, -\, x^2)\, dx\, =\, \dfrac{125}{3}$$

But I get stuck when I try to evaluate it:

. . . . .$$\displaystyle \displaystyle 3\int_0^b\, (25\, -\, x^2)\, =\, 125$$

. . . . .$$\displaystyle \displaystyle 3\left[\, 25x\, -\, \frac{1}{3}x^3\right]_0^b\, =\, 125$$

. . . . .$$\displaystyle \displaystyle 75b\, -\, b^3\, =\, 125$$
You're using a line x=b, not y=b, to divide the region!

Try again. There are a couple ways you might do this.

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#### Seed5813

##### New member
You're using a line x=b, not y=b, to divide the region!

Try again. There are a couple ways you might do this.
Y= 25-x^2
y-25=x^2
sqrt(25-y)=x

Int_0^25 (25-y)^(1/2)dy
U=25-y
du=-dy
-int_0^25(u^(1/2))du
-[(2/3)u^(3/2)]_0^25
-[(2/3)(25-y)^(3/2)]_0^25
-[-(2/3)(25)^(3/2)]
[(2/3)(125)]
(250/3)

Also how do I know when to use x and when to use y?

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#### Seed5813

##### New member
Y= 25-x^2
y-25=x^2
sqrt(25-y)=x

Int_0^25 (25-y)^(1/2)dy
U=25-y
du=-dy
-int_0^25(u^(1/2))du
-[(2/3)u^(3/2)]_0^25
-[(2/3)(25-y)^(3/2)]_0^25
-[-(2/3)(25)^(3/2)]
[(2/3)(125)]
(250/3)

Also how do I know when to use x and when to use y?

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^wrong btw

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#### Seed5813

##### New member
Sorry here's the whole thing:

Y= 25-x^2
y-25=x^2
sqrt(25-y)=x

Int_0^25 (25-y)^(1/2)dy
U=25-y
du=-dy
-int_0^25(u^(1/2))du
-[(2/3)u^(3/2)]_0^25
-[(2/3)(25-y)^(3/2)]_0^25
-[-(2/3)(25)^(3/2)]
[(2/3)(125)]
(250/3)

Int_0^b ((25-y)^(1/2)dy=(250/3)
U=25-y
Du=-dy
-int_0^b(u^(1/2))du=(250/3)
-[(2/3)(25-y)^(3/2)]_0^b=(250/3)
-[(2/3)(25-b)^(3/2) - (250/3)]=(250/3)
-(2/3)(25-b)^(3/2) + (250/3) = (250/3)
-(2/3)(25-b)^(3/2) = 0
(25-b)^(3/2)=0
25-b = 0
B=-25

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#### Seed5813

##### New member
You're using a line x=b, not y=b, to divide the region!

Try again. There are a couple ways you might do this.
Found the problem

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#### Dr.Peterson

##### Elite Member
Sorry here's the whole thing:

Y= 25-x^2
y-25=x^2
sqrt(25-y)=x

Int_0^25 (25-y)^(1/2)dy
U=25-y
du=-dy
-int_0^25(u^(1/2))du
-[(2/3)u^(3/2)]_0^25
-[(2/3)(25-y)^(3/2)]_0^25
-[-(2/3)(25)^(3/2)]
[(2/3)(125)]
(250/3)

Int_0^b ((25-y)^(1/2)dy=(250/3)
U=25-y
Du=-dy
-int_0^b(u^(1/2))du=(250/3)
-[(2/3)(25-y)^(3/2)]_0^b=(250/3)
-[(2/3)(25-b)^(3/2) - (250/3)]=(250/3)
-(2/3)(25-b)^(3/2) + (250/3) = (250/3)
-(2/3)(25-b)^(3/2) = 0
(25-b)^(3/2)=0
25-b = 0
B=-25
You're working only with the right half of the region, but that doesn't affect the answer.

The trouble here is that you didn't take half of the area (125/3), so your answer turned out to be actually b=+25, which is the whole region.

I presume you realized that and got the correct answer.

There's a formula for the area of a segment of a parabola: 2bh/3. But you were, of course, expected to do what you did.