# Homework Help: Integration/Solving A Constant To Divide An Area

1. May 23, 2010

### Lancelot59

With this particular problem you are given the functions y=x$$^{2}$$, y=4, and y=b.

The object is to solve for a value of b that will split the area between y=4 and y=x$$^{2}$$ in half.
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP263419aghg2f97i2944e0000580a2hgi735b957c?MSPStoreType=image/gif&s=42&w=399&h=201 [Broken]

The approach I took was to include y=b in the integral, and then subtract the top half from the bottom half and equate that to zero.

$$\int_{-2}^{2} {(4-x^2-b)}\ = \int_{-2}^{2} {(b-x^2)}$$

All I did then was solve the integrals, and isolate b to get an answer of 4. However the answer is 4^2/3. I'm also stuck on a similar problem where I need to solve for a constant where two functions have to enclose a specific area.

Last edited by a moderator: May 4, 2017
2. May 23, 2010

### gabbagabbahey

Your integral for the top half is incorrect. Using the "vertical slices" method, the lower limit on $y$ is $y=b$ when $-\sqrt{b}<x<\sqrt{b}$, and $y=x^2$ when $|x|\geq\sqrt{b}$. So, you might want to either use horizontal slices for the top half, or not bother computing that integral at all and instead compute the area between $y=x^2$ and $y=4$ and set that equal to twice your bottom half.

3. May 23, 2010

### Lancelot59

Horizontal slices sounds like a better idea. I did come up with the +-root b being the limits, but I decided to go with negative two and two instead...apparently that didn't work. I guess just integrating for one half only works out nicely. Then I can just go from zero to 4.

Last edited: May 23, 2010
4. May 23, 2010

### Lancelot59

Okay, so I did it with respect to Y instead:

$$\int_{0}^{b} {\sqrt{y}}\ = \int_{0}^{4} {\sqrt{y}}\ - \int_{0}^{b} {\sqrt{y}}$$

then got here:

$$[\sqrt{b} - \sqrt{0}] = [\sqrt{4} - \sqrt{0}] - [\sqrt{b} - \sqrt{0}]$$

then:
$$2\sqrt{b}= \sqrt{4}$$

$$2\sqrt{b}= 2$$

$$b= 1$$

Things aren't going very well.

Last edited: May 23, 2010
5. May 23, 2010

### happyg1

I found the desired area to be 8/3 (I just chopped it in half and did 0 to 2). Then integrated $$\frac{8}{3}=\int_{0}^{\sqrt b} b-x^2 dx$$
Solved for b...

6. May 23, 2010

### Lancelot59

You mean
$$\int_{0}^{\sqrt b} { [b-x^2]} dx$$

? I'll give it a try.

7. May 23, 2010

### happyg1

Yes, My tex skills are a little rusty. i edited it and fixed it..

8. May 23, 2010

### Lancelot59

Mine are horrible. I just figured it out within the last hour to ask this question. :p

Also, that doesn't work. The answer is supposed to be the cube root of 16...I am confused.

9. May 23, 2010

### happyg1

set it equal to 8/3! It works!

10. May 23, 2010

### Lancelot59

And the 8/3 is half the parabola from 0 to 2? Attempt # 4ish...

11. May 23, 2010

### Lancelot59

Well, here's what I did:

$$\frac{8}{3}=\int_{0}^{\sqrt b} b-x^2 dx$$

$$\frac{8}{3}=[b-{\sqrt b}^2] - [b-{\sqrt 0}^2]$$

$$\frac{8}{3}= b-b-b$$

$$\frac{8}{3}= -b$$

I'm really confused now.

12. May 23, 2010

### happyg1

$$\frac{8}{3}=\int_{0}^{\sqrt b} b-x^2 dx =bx-\frac{1}{3}x^3$$ from 0 to $$\sqrt b$$

13. May 23, 2010

### Cyosis

I suggest you review basic integration. You're just plugging the numbers into the integrand instead of integrating.

14. May 23, 2010

### Lancelot59

...Fail. I've been doing that a lot recently. I don't know why.