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

The integral from 0 to 9-z^2-y^2 of z^2 is ((-z^2+9-y^2)^3)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 9-z^2-y^2 of z^2

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

\frac{( - z ^{2} + 9 - y ^{2} ) ^{3}}{3}

Solution steps

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

Apply the Power Rule

: [\frac{z ^{3}}{3}]_{0}^{9 - y^{2} - z^{2}}

= [\frac{z ^{3}}{3}]_{0}^{9 - y^{2} - z^{2}}

Compute the boundaries:

\frac{1}{3} (- z^{2} + 9 - y^{2})^{3}

= \frac{1}{3} (- z^{2} + 9 - y^{2})^{3}

Simplify

= \frac{( - z ^{2} + 9 - y ^{2} ) ^{3}}{3}

\int_{1}^{6} \frac{ln ( x )}{x} d x

\int_{0}^{2} y^{2} sin (π y^{3}) d y

\int_{0.1}^{0.2} \frac{1}{x ^{2}} d x

\int_{0}^{\frac{π}{3}} 2 sec^{3} (x) d x

\int_{1}^{5} \frac{1}{( x - 2 ) ^{\frac{1}{3}}} d x

What is the integral from 0 to 9-z^2-y^2 of z^2 ?
The integral from 0 to 9-z^2-y^2 of z^2 is ((-z^2+9-y^2)^3)/3