What is the integral from z^2+y^2 to 1 of z ?

The integral from z^2+y^2 to 1 of z is (1-y^4-2y^2z^2-z^4)/2

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integral from z^2+y^2 to 1 of z

\int_{z^{2} + y^{2}}^{1} z d z

\frac{1 - y ^{4} - 2 y ^{2} z ^{2} - z ^{4}}{2}

Solution steps

\int_{z^{2} + y^{2}}^{1} z d z

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{y^{2} + z^{2}}^{1}

= [\frac{z ^{2}}{2}]_{y^{2} + z^{2}}^{1}

Compute the boundaries:

\frac{1}{2} - \frac{( y ^{2} + z ^{2} ) ^{2}}{2}

= \frac{1}{2} - \frac{( y ^{2} + z ^{2} ) ^{2}}{2}

Simplify

= \frac{1 - y ^{4} - 2 y ^{2} z ^{2} - z ^{4}}{2}

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What is the integral from z^2+y^2 to 1 of z ?
The integral from z^2+y^2 to 1 of z is (1-y^4-2y^2z^2-z^4)/2