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

The integral from 0 to sqrt(9-y^2 of)z is (-y^2+9)/2

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integral from 0 to sqrt(9-y^2 of)z

\int_{0}^{9 - y^{2}} z d z

\frac{- y ^{2} + 9}{2}

Solution steps

\int_{0}^{9 - y^{2}} z d z

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{0}^{9 - y^{2}}

= [\frac{z ^{2}}{2}]_{0}^{9 - y^{2}}

Compute the boundaries:

\frac{- y ^{2} + 9}{2}

= \frac{- y ^{2} + 9}{2}

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What is the integral from 0 to sqrt(9-y^2 of)z ?
The integral from 0 to sqrt(9-y^2 of)z is (-y^2+9)/2