What is the integral from y^2 to 8-2x^2-y^2 of 10xz ?

The integral from y^2 to 8-2x^2-y^2 of 10xz is 20x(x^4+x^2y^2-8x^2+16-4y^2)

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integral from y^2 to 8-2x^2-y^2 of 10xz

\int_{y^{2}}^{8 - 2 x^{2} - y^{2}} 10 x z d z

20 x (x^{4} + x^{2} y^{2} - 8 x^{2} + 16 - 4 y^{2})

Solution steps

\int_{y^{2}}^{8 - 2 x^{2} - y^{2}} 10 x z d z

Take the constant out:

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

= 10 x \cdot \int_{y^{2}}^{8 - y^{2} - 2 x^{2}} z d z

Apply the Power Rule

= 10 x [\frac{z ^{2}}{2}]_{y^{2}}^{8 - y^{2} - 2 x^{2}}

Compute the boundaries:

\frac{( 8 - y ^{2} - 2 x ^{2} ) ^{2}}{2} - \frac{y ^{4}}{2}

= 10 x (\frac{( 8 - y ^{2} - 2 x ^{2} ) ^{2}}{2} - \frac{y ^{4}}{2})

Simplify

= 20 x (x^{4} + x^{2} y^{2} - 8 x^{2} + 16 - 4 y^{2})

What is the integral from y^2 to 8-2x^2-y^2 of 10xz ?
The integral from y^2 to 8-2x^2-y^2 of 10xz is 20x(x^4+x^2y^2-8x^2+16-4y^2)