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

The integral from 0 to 2-x^2-y^2 of (xye^z) is xy(e^{2-x^2-y^2}-1)

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integral from 0 to 2-x^2-y^2 of (xye^z)

\int_{0}^{2 - x^{2} - y^{2}} (x y e^{z}) d z

x y (e^{2 - x^{2} - y^{2}} - 1)

Solution steps

\int_{0}^{2 - x^{2} - y^{2}} x y e^{z} d z

Take the constant out:

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

= x y \cdot \int_{0}^{2 - x^{2} - y^{2}} e^{z} d z

Use the common integral:

\int e^{z} d z = e^{z}

= x y [e^{z}]_{0}^{2 - x^{2} - y^{2}}

Compute the boundaries:

e^{2 - x^{2} - y^{2}} - 1

= x y (e^{2 - x^{2} - y^{2}} - 1)

\int_{- h}^{h} (y^{2} - h^{2}) d y

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What is the integral from 0 to 2-x^2-y^2 of (xye^z) ?
The integral from 0 to 2-x^2-y^2 of (xye^z) is xy(e^{2-x^2-y^2}-1)