What is the integral from 0 to 3 of sqrt(9-r^2)r ?

The integral from 0 to 3 of sqrt(9-r^2)r is 9

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integral from 0 to 3 of sqrt(9-r^2)r

\int_{0}^{3} 9 - r^{2} r d r

9

Solution steps

\int_{0}^{3} 9 - r^{2} r d r

Apply u-substitution

: \int_{9}^{0} - \frac{u}{2} d u

= \int_{9}^{0} - \frac{u}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{9} - \frac{u}{2} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{0}^{9} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

Compute the boundaries:

18

= \frac{1}{2} \cdot 18

Simplify

= 9

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What is the integral from 0 to 3 of sqrt(9-r^2)r ?
The integral from 0 to 3 of sqrt(9-r^2)r is 9