What is the integral from 0 to r of sqrt(r^2-y^2) ?

The integral from 0 to r of sqrt(r^2-y^2) is pi/4 r^2

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integral from 0 to r of sqrt(r^2-y^2)

\int_{0}^{r} r^{2} - y^{2} d y

\frac{π}{4} r^{2}

Solution steps

\int_{0}^{r} r^{2} - y^{2} d y

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} r^{2} cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} r^{2} cos^{2} (u) d u

Take the constant out:

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

= r^{2} \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= r^{2} \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= r^{2} \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 u) d u

Apply the Sum Rule:

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

= r^{2} \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= r^{2} \frac{1}{2} (\frac{π}{2} + 0)

Simplify

r^{2} \frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{4} r^{2}

= \frac{π}{4} r^{2}

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\frac{d y}{d x} = x^{3} y^{3}

\frac{\partial}{\partial x} (x arcsin (\frac{y}{x}))

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What is the integral from 0 to r of sqrt(r^2-y^2) ?
The integral from 0 to r of sqrt(r^2-y^2) is pi/4 r^2