What is the integral from 0 to 8 of 2piy(64-y^2) ?

The integral from 0 to 8 of 2piy(64-y^2) is 2048pi

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integral from 0 to 8 of 2piy(64-y^2)

\int_{0}^{8} 2 π y (64 - y^{2}) d y

2048 π

Decimal Notation

6433.98175 \dots

Solution steps

\int_{0}^{8} 2 π y (64 - y^{2}) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π \cdot \int_{0}^{8} y (64 - y^{2}) d y

Expand

y (64 - y^{2}) : 64 y - y^{3}

= 2 π \cdot \int_{0}^{8} 64 y - y^{3} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 π (\int_{0}^{8} 64 y d y - \int_{0}^{8} y^{3} d y)

\int_{0}^{8} 64 y d y = 2048

\int_{0}^{8} y^{3} d y = 1024

= 2 π (2048 - 1024)

Simplify

= 2048 π

\frac{4}{s ^{2} ( s ^{2} + 16 )}

y = tan (x), at x = π

\int x^{4} (5 - x^{2}) d x

\frac{\partial}{\partial y} (- \frac{x}{( x + y ) ^{2}})

\frac{d}{d x} (\frac{x}{2 + c x})

What is the integral from 0 to 8 of 2piy(64-y^2) ?
The integral from 0 to 8 of 2piy(64-y^2) is 2048pi