Question

Calculus problem! please help! (picture included)

Answer 1

f(x)=4/3 pi r^3

Answer 2

g(x)=4/3 pi r^3

Answer 3

is that the answer? can you show me each steps?:)

Answer 4

No, that's not the answer..

Answer 5

You are looking for a piecewise

Answer 6

find the volume of glass of wine without the cherry

Answer 7

as a function with f(h)=V

Answer 8

then, find how much the cherry displaces the volume

Answer 9

can you write out the steps please?

Answer 10

That would require me to look at parametrization, and I'm too lazy. Instead, @KingGeorge

Answer 11

also, George, can you call other peoples?

Answer 12

Interesting and rather difficult calc problem to be had here @satellite73 @saifoo.khan @amistre64 @myininaya @Hero @AccessDenied @experimentX

Answer 13

by the way, what calc are you taking?

Answer 14

Calc 2, i dont get these problems and i have a few more. ahahaha..

Answer 15

Perhaps the best way to think about this would be as a rotation around the y-axis. If you centered the half-circle of the cup at (0, 0), you could find the volume using rotations.

Answer 16

If it helps you think about it, you could just flip the diagram on its side and rotate around the x-axis instead.

Answer 17

I would appreciate it if you could show me the steps:)

Answer 18

If we flip it so its around the x-axis instead, the big circle would be given by equation \[x^2+y^2=4^2\]And the equation for the circle of the cherry would be \[(x-3)^2+y^2 =1^2\]

Answer 19

Now we need to look at the bounds for the piecewise function we're bound to have. If the liquid is completely covering the cherry, we know that \(0\leq x\leq2\), and if the liquid is only partially covering the cherry, \(2 < x\leq 4\).

So those are our two bounds for the upcoming piecewise equation.

Answer 20

Let's start with the easy one. Let \(0\leq x\leq2\). Then the volume of the larger sphere that's filled with liquid would be given by \[\pi \int\limits_h^4 \sqrt{16-x^2}^2\quad dx\]In other terms, \[\pi \int\limits\limits_h^4 16-x^2\quad dx\]Can you evaluate this integral yourself?

Answer 21

yes thank you so much!

Answer 22

There's still more though! That integral is neglecting the cherry that's still in the glass. Fortunately, the cherry is completely covered so we can just subtract by the volume which is \(4\pi/3\). So that's the first piecewise part of the problem.

Answer 23

See, it *was* piecewise! YAHOOO

Answer 24

oh :O ... so I got (h^3/3 -16h + 128/3)π then what do you do..?

Answer 25

IT WAS PIECEWISE, I AM ALWAYYSSS correct...

Answer 26

Now for the second part. This is where the cherry is only partly covered so \(2 < x\leq4\). Also, we'll have to subtract the volume of the cherry before we integrate. Knowing this, we can integrate as follows\[\pi \int\limits_h^4 \sqrt{16-x^2}^2\quad dx - \pi \int\limits_h^4 \sqrt{1-(x-3)^2}^2 \quad dx\]

Answer 27

so it would be π(2h^3/3 -3h^2 -8h + 112/3) and then?

Answer 28

And btw, for the first integral, I'm getting \(-h^3/3 -16h+128/3\)

Answer 29

Times \(\pi\) of course.

Answer 30

the first one h^3/3 is positive.

Answer 31

Oh yes. Of course. Silly me.

Answer 32

are you playing in R^2? instead of R^3?

Answer 33

im getting so confused ... :O

Answer 34

We're rotating about the axis instead of using triple integrals.

Answer 35

And I'm not getting π(2h^3/3 -3h^2 -8h + 112/3) for the final solution. Rather, I'm getting \(V=\pi(48-24 h+3 h^2)\)

Answer 36

cherry eq = (x-1)^2+y^2 = 2^2

glass eq = (x-4)^2 +y^2 = 4^2

right?

Answer 37

amistre's way will also work, and might actually be an easier way. I've got to leave for a bit now, but good luck.

PS. radius of the cherry is 1 not 2.