# Thread: Help me with calculus question, due at 12 oclock eastern time!

1. ## Help me with calculus question, due at 12 oclock eastern time!

Hi everyone, new to the forums here. I need help on this question ASAP, it's due in less than 15 minutes and I have to submit it online. Here's the question...

Find the number b such that the line y = b divides the region bounded by the curves y = x2 and y = 9 into two regions with equal area. (Round your answer to the nearest hundredth.)

Help asap, thank you!
Nate

2. Originally Posted by flairknocks
Hi everyone, new to the forums here. I need help on this question ASAP, it's due in less than 15 minutes and I have to submit it online. Here's the question...

Find the number b such that the line y = b divides the region bounded by the curves y = x2 and y = 9 into two regions with equal area. (Round your answer to the nearest hundredth.)

Help asap, thank you!
Nate
Why'd you wait so late?

3. I've been busy with school band, other homework, etc.

4. Originally Posted by flairknocks
I've been busy with school band, other homework, etc.
So...What have you done to try and takle this thing?

5. I've found the area between the lines, but I don't know how to divide it. I thought it was zero, but then that's x=0. I need y=0, right?

6. Originally Posted by flairknocks
I've found the area between the lines, but I don't know how to divide it. I thought it was zero, but then that's x=0. I need y=0, right?
To find the area

$\int_{-3}^{3}(9-x^2)dx$

Then, flip your graph on its side and find the average value...That is

$\frac{1}{9}\int_0^9\sqrt{y}dy$

7. Originally Posted by flairknocks
Hi everyone, new to the forums here. I need help on this question ASAP, it's due in less than 15 minutes and I have to submit it online. Here's the question...

Find the number b such that the line y = b divides the region bounded by the curves y = x2 and y = 9 into two regions with equal area. (Round your answer to the nearest hundredth.)

Help asap, thank you!
Nate
Area = $\int_{-3}^3 9-x^2\,dx = 36$. So half the area is $18$. You want to find $y^2$ such that $\int_{-y}^y y^2-x^2\,dx = 18$.

Integral equals $\left[xy^2-\frac{x^3}{3}\right]_{-y}^y = \frac{4y^3}{3} = 18$.

Therefore $y= \left(\frac{27}{2}\right)^{1/3}$ but remember you want $y^2$ so the final answer is $\boxed{y^2= \left(\frac{27}{2}\right)^{2/3} = 5.67}$