We will lay the cask on its side to make the algebra easier:

Parabola with vertex at `(0, 40)` and passing through `(50, 30)`.

We need to find the equation of a parabola with vertex at `(0, 40)` and passing through `(50, 30)`.

We use the formula:

(

x−h)^{2}= 4a(y−k)

Now (*h*,* k*) is (0, 40) so we have: *x*^{2} = 4*a*(*y *− 40) and the parabola passes through (50, 30), so

(50)

^{2}= 4a(30 − 40)

2500 = 4

a(−10) and this gives 4a= −250

So the equation of the side of the barrel is

x^{2}= -250(y− 40), that is,

`y=-(x^2)/(250)+40`

We need to find the volume of the cask which is generated when we rotate this parabola between *x* = -50 and *x* = 50 around the *x*-axis.

Parabolic area rotated around the `x`-axis producing our wine cask.

We now apply the formula for the volume of a solid of revolution:

`"Vol"=pi int_a^by^2\ dx`

`=pi int_-50^50(-(x^2)/(250)+40)^2 dx`

`=pi int_-50^50((x^4)/(62500) - (80x^2)/(250)+1600) dx`

`=pi[(x^5)/(312500)-(80x^3)/(750)+1600x]_-50^50`

Now, since

(−50)

^{5}= −50^{5},(−50)

^{3}= −50^{3}, and(−50) = −50,

we can reduce the amount of writing somewhat and put:

`text[Vol] = 2pi[((50)^5)/(312500)-(80(50)^3)/(750)+1600(50)]`

`=425162\ text[cm]^3`

`=425.2\ text[L]`

So the wine cask will hold `425.2\ "L"`.

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