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}

\int \frac{8 x ^{2}}{( x ^{3} + 2 ) ^{3}} d x

\int \frac{( ln ( x ) ) ^{4}}{x} d x

\int_{1}^{e} x^{2} ln (x) d x

\frac{y ^{4}}{8} + \frac{1}{4 y ^{2}}

5 y^{^{''}} + 26 y = 0

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