# Thread: The volume of a solid between two curves

1. ## The volume of a solid between two curves

I need to find the volume of a solid between two curves. However, I am struggling to find the points of intersection of these two curves. I am making them equal to each other but I am not sure how to solve the equation to find the two points. The two curves are:

$
y = 10x - x^2
$

and
$
y = e^x-1
$

Any help is greatly appreciated!

2. Originally Posted by hasanbalkan
I need to find the volume of a solid between two curves. However, I am struggling to find the points of intersection of these two curves. I am making them equal to each other but I am not sure how to solve the equation to find the two points. The two curves are:

$
y = 10x - x^2
$

and
$
y = e^x-1
$

Any help is greatly appreciated!
Two things:
1) You are looking for an area, not a volume.

2) Your approach is correct, but there is no algebraic way to solve this system. You'll have to solve it numerically. However note that there is a solution for x = 0. Off the top of my head I can't find another nice solution.

-Dan

3. Thank you for the reply. Actually, my book says that I need to find the volume - I double checked. I found a formula for the volume. The only thing I need to tackle is to find the intersection points. How do I apply a numerical method of solving it?

4. You need to know the line about which it is rotated. Without it, we cannot find a reasonable value for any variable in the formula.

5. Originally Posted by Aryth
You need to know the line about which it is rotated. Without it, we cannot find a reasonable value for any variable in the formula.
It is rotated around the y-axis.

6. So, first, we have to find the equations for the inner and outer radii that I will denote as:

$Outer \ Radius = R(y)$
$Inner \ Radius = r(y)$

So, if you graph it and look at the two graphs, you can easily see that the outer radius is just the x-value of the outer function, so all we have to do is solve the outer function for x (the $e^x$ function):

$y = e^x - 1$
$R(y) = ln(y + 1)$

Now, if you look at the inner radius, it's just the x value of the inner function (the parabola), so all you do is solve for x:

$y = 10x - x^2$

$-x^2 + 10x - y = 0$

Now we must use the quadratic formula (But we are only going to use the part with the positive radical since that makes the half of the parabola we are concerned with):

$x = \frac{-10 + \sqrt{100 - 4(-1)(-y)}}{-2}$

$r(y) = \frac{10 - \sqrt{100 - 4y}}{2}$

Just check a few problems:

for x = 1:

$y = 10 - 1 = 9$

for y = 9:

$x = \frac{10 - \sqrt{64}}{2}$
$x = 1$

Ok, now we have to find the limits by setting the equations together... To make this shorter I used my graph and discovered that the limits are:

$x \approx 3.11$
$x = 0$

So, now we set up our integral:

$V = \pi \int_{0}^{3.11}[R(y)^2 - r(y)^2]dy$

$V = \pi \int_{0}^{3.11}[(2ln(y + 1) - \left(\frac{10 - \sqrt{100 + 4y}}{2}\right)^2]dy$

Where $r(y)^2$ simplifies to:

$\left(\frac{(10 - \sqrt{100 + 4y})^2}{4}\right)$

$= \left(\frac{100 - 20\sqrt{100 + 4y} + 4y}{4}\right)$

$= 25 - 5\sqrt{100 + 4y} + y$

Plug that in:

$V = \pi \int_{0}^{3.11}[2ln(y + 1) - (25 - 5\sqrt{100 + 4y} + y)]dy$

I believe you can solve from there or use another method if you'd like. This is the washer method by the way.

7. Since they intersect at x= $\frac{1944}{625}$, then if you use shells:

$2{\pi}\int_{0}^{\frac{1944}{625}}x\left[(10x-x^{2})-(e^{x}-1)\right]dx$

8. If you want to use washers, then:

${\pi}\int_{0}^{\frac{2143}{100}}\left[(ln(y+1))^{2}-(-(\sqrt{25-y}-5))^{2}\right]dy$

$={\pi}\int_{0}^{\frac{2143}{100}}\left[(ln(y+1))^{2}+10\sqrt{25-y}+y-50\right]dy$

You will get the same result as with shells.