1. ## Another application Integral

I was just looking over my notes and I do not understand this question entirely

A cone with height 10ft and top radius 5ft is filled with oil which weighs 57lb/ft^3. find work done in pumping out the oil.

I know that the interval being sliced is [0,10] because the oil fills the entire tank.

I also understand why the the distance is (10-y)

What I do not understand according to my notes is where (y^2) comes from. Or where the radius comes into play?

If it is coming from the formula for a cone 1/3pi*r^2 then why does my notes say the final answer is

$\frac{57{\pi}}{4}\int_{0}^{10}(y^2)(10-y)dy$

I thought it should be $\frac{57{\pi}}{3}$

2. Originally Posted by gammaman
i was just looking over my notes and i do not understand this question entirely

a cone with height 10ft and top radius 5ft is filled with oil which weighs 57lb/ft^3. Find work done in pumping out the oil.

I know that the interval being sliced is [0,10] because the oil fills the entire tank.

I also understand why the the distance is (10-y)

what i do not understand according to my notes is where (y^2) comes from. Or where the radius comes into play?

If it is coming from the formula for a cone 1/3pi*r^2 then why does my notes say the final answer is

$\frac{57{\pi}}{4}\int_{0}^{10}(y^2)(10-y)dy$

i thought it should be $\frac{57{\pi}}{3}$

See the attachment for the similar triangles

from this we know that the radius is

$r=\frac{1}{2}(10-y)=\frac{10-y}{2}$ so each little volume chunk is a cyllindar of height dy and and above radius

$dV=\pi \left( \frac{10-y}{2}\right)^2dy$

so now Force will be density times volume

$dF=\frac{57\pi}{4}(10-y)^2dy$

Since we will move each dy slice y units we get

$\int_{0}^{10}y \cdot \frac{57\pi}{4}(10-y)^2dy=\frac{57\pi}{4}\int_{0}^{10}y(10-y)^2dy$

3. why are you squaring (10-y) and not y?

4. Originally Posted by gammaman
why are you squaring (10-y) and not y?
If you look at the diagram you will see why. We have the roles of y and y-10 exchanged in our diagrams. This is unimportant the integrals are equivelent. Basically I didn't notice that you have your coordinate system starting at the bottom of the cone where mine is at the top.

if you let $u=10-y \iff y=10-u \implies dy=-du$

$\int_{0}^{10}y(10-y)^2dy=\int_{10}^{0}(10-u)(u)^2(-du)=\int_{0}^{10}u^2(10-u)du$

Remember the way you set up alot of these is not unique.

5. Ok so the formula's do not change. The only thing that would change would be the Di. If I adding fluid it is (y + something). If I am taking fluid out it is
(Y-something).

By the way, I love your avatar, is that from Oblivion?

6. So let me get this straight. If the height were 12 and the radius were 3, I would have

$\frac{Yi}{12}=\frac{Ri}{3}$

Then $\frac{3Yi}{12}=\frac{12Ri}{12}$

So then we would have (.25)^2 = 1/8?

so then it would just be $\frac{57{\pi}}{8}\int_{0}^{12}(y^2)(12-y)dy$

7. Originally Posted by gammaman
So let me get this straight. If the height were 12 and the radius were 3, I would have

$\frac{Yi}{12}=\frac{Ri}{3}$

Then $\frac{3Yi}{12}=\frac{12Ri}{12}$

So then we would have (.25)^2 = 1/8?

so then it would just be $\frac{57{\pi}}{8}\int_{0}^{12}(y^2)(12-y)dy$
Your logic is correct but $(.25)^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}$

8. so then it would be $\frac{57{\pi}}{16}?$

9. Originally Posted by gammaman
so then it would be $\frac{57{\pi}}{16}?$

You got it!

10. One last question, what would the "setup" be for a cylinder?

And I am still dying to know where your avatar comes from.

11. Originally Posted by gammaman
One last question, what would the "setup" be for a cylinder?

And I am still dying to know where your avatar comes from.
Sorry my wife found it for me and it is supposed to be Raistlin Mejere from Dragon lance. I can ask her where she got it when she gets home from school. off topic I just played my first oblivion game on PS3 and it was awsome. On topic

We are using the height as dy so on the small scale each slice is approximated by a cylindar. Think of it as we are cutting little circular disks of height dy