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

2550-25-5010203040-10-20-30-40xy

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: x2 = 4a(y − 40) and the parabola passes through (50, 30), so

(50)2 = 4a(30 − 40)

2500 = 4a(−10) and this gives 4a = −250

So the equation of the side of the barrel is

x2 = -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.

2550-25-502040-20-40xy

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 = −505,

(−50)3 = −503, 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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