What is the integral of sqrt(4-x^2-y^2) ?

The integral of sqrt(4-x^2-y^2) is

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integral of sqrt(4-x^2-y^2)

\int 4 - x^{2} - y^{2} d x

2 arcsin (\frac{x}{4 - y ^{2}}) + sin (2 arcsin (\frac{x}{4 - y ^{2}})) - \frac{1}{2} y^{2} arcsin (\frac{x}{4 - y ^{2}}) - \frac{1}{4} y^{2} sin (2 arcsin (\frac{x}{4 - y ^{2}})) + C

Solution steps

\int 4 - x^{2} - y^{2} d x

Apply Trigonometric Substitution

: \int cos^{2} (u) (4 - y^{2}) d u

= \int cos^{2} (u) (4 - y^{2}) d u

Take the constant out:

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

= (4 - y^{2}) \cdot \int cos^{2} (u) d u

\int cos^{2} (u) d u = \frac{1}{2} (u + \frac{1}{2} sin (2 u))

= (4 - y^{2}) \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{4 - y ^{2}} x)

= (4 - y^{2}) \frac{1}{2} (arcsin (\frac{1}{4 - y ^{2}} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{4 - y ^{2}} x)))

Simplify

(4 - y^{2}) \frac{1}{2} (arcsin (\frac{1}{4 - y ^{2}} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{4 - y ^{2}} x))) : 2 arcsin (\frac{x}{4 - y ^{2}}) + sin (2 arcsin (\frac{x}{4 - y ^{2}})) - \frac{1}{2} y^{2} arcsin (\frac{x}{4 - y ^{2}}) - \frac{1}{4} y^{2} sin (2 arcsin (\frac{x}{4 - y ^{2}}))

= 2 arcsin (\frac{x}{4 - y ^{2}}) + sin (2 arcsin (\frac{x}{4 - y ^{2}})) - \frac{1}{2} y^{2} arcsin (\frac{x}{4 - y ^{2}}) - \frac{1}{4} y^{2} sin (2 arcsin (\frac{x}{4 - y ^{2}}))

Add a constant to the solution

= 2 arcsin (\frac{x}{4 - y ^{2}}) + sin (2 arcsin (\frac{x}{4 - y ^{2}})) - \frac{1}{2} y^{2} arcsin (\frac{x}{4 - y ^{2}}) - \frac{1}{4} y^{2} sin (2 arcsin (\frac{x}{4 - y ^{2}})) + C

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