What is the integral of (1-x^2-y^2) ?

The integral of (1-x^2-y^2) is y-x^2y-(y^3)/3+C

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integral of (1-x^2-y^2)

\int (1 - x^{2} - y^{2}) d y

y - x^{2} y - \frac{y ^{3}}{3} + C

Solution steps

\int 1 - x^{2} - y^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d y - \int x^{2} d y - \int y^{2} d y

\int 1 d y = y

\int x^{2} d y = x^{2} y

\int y^{2} d y = \frac{y ^{3}}{3}

= y - x^{2} y - \frac{y ^{3}}{3}

Add a constant to the solution

= y - x^{2} y - \frac{y ^{3}}{3} + C

x \to - 3 - lim (\frac{∣ x - 3 ∣}{x - 3})

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What is the integral of (1-x^2-y^2) ?
The integral of (1-x^2-y^2) is y-x^2y-(y^3)/3+C