# A ball of radius 11 has a round hole of radius 5 drilled through its center.Find the volume of the resulting solid. Find it using integrals. Find the volume the way that is used in calculus 2

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You should use the equation of the circle such that:

`x^2 + y^2 = 11^2 => y^2 = 121 - x^2 => y = +-sqrt(121 - x^2)`

You need to know that rotating the circle having the equation `x^2 + y^2 = 11^2` about x axis yields the volume of the ball.

Hence, you should use the following formula of volume such that:

`V = pi*int_a^b y^2 dx`

You should solve the following system of equations to find the limits of integration such that:

`{(y = 5),(y = +-sqrt(121 - x^2)):} => 5 = +-sqrt(121 - x^2)`

`25 = 121 - x^2 => x^2 = 121 - 25 = 96 => x_(1,2) = +-sqrt96`

`V = pi*int_-sqrt96^sqrt96 (sqrt(121 - x^2))^2 dx`

Notice that the function `(sqrt(121 - x^2))^2` is even, hence, you should use the following identity such that:

`int_-a^a f(x)dx = 2int_0^a f(x)dx`

`V = 2pi*int_0^sqrt96(121 - x^2) dx`

`V = 2pi(121x - x^3/3)|_0^sqrt96`

`V = 2pi(121*sqrt96 - sqrt96/3) => V = 2pi*362sqrt96/3`

`V = 2896pi*sqrt6/3`

**Hence, evaluating the volume of resulting solid, under the given conditions, yields `V = 965.3*pi*sqrt6` .**