What is the integral from z^2+y^2 to 36 of 6z ?

The integral from z^2+y^2 to 36 of 6z is 6(648-((y^2+z^2)^2)/2)

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integral from z^2+y^2 to 36 of 6z

\int_{z^{2} + y^{2}}^{36} 6 z d z

6 (648 - \frac{( y ^{2} + z ^{2} ) ^{2}}{2})

Solution steps

\int_{z^{2} + y^{2}}^{36} 6 z d z

Take the constant out:

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

= 6 \cdot \int_{y^{2} + z^{2}}^{36} z d z

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{y^{2} + z^{2}}^{36}

= 6 [\frac{z ^{2}}{2}]_{y^{2} + z^{2}}^{36}

Compute the boundaries:

648 - \frac{( y ^{2} + z ^{2} ) ^{2}}{2}

= 6 (648 - \frac{( y ^{2} + z ^{2} ) ^{2}}{2})

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What is the integral from z^2+y^2 to 36 of 6z ?
The integral from z^2+y^2 to 36 of 6z is 6(648-((y^2+z^2)^2)/2)