What is the integral from 0 to 6r of sqrt(36r^2-y^2) ?

The integral from 0 to 6r of sqrt(36r^2-y^2) is 9pir^2

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integral from 0 to 6r of sqrt(36r^2-y^2)

\int_{0}^{6 r} 36 r^{2} - y^{2} d y

9 π r^{2}

Solution steps

\int_{0}^{6 r} 36 r^{2} - y^{2} d y

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 36 r^{2} cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 36 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

= 36 r^{2} \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 36 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

= 36 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

= 36 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

= 36 r^{2} \frac{1}{2} (\frac{π}{2} + 0)

Simplify

36 r^{2} \frac{1}{2} (\frac{π}{2} + 0) : 9 π r^{2}

= 9 π r^{2}

x \to \infty lim (ln (6 x))

y^{3} - (2 x + 6) + 3 x y^{2} y^{^{'}} = 0

g (x) = 13 x

x \to - 1 + lim (2 - x)

x \to 6 + lim (\frac{2}{x - 6})

What is the integral from 0 to 6r of sqrt(36r^2-y^2) ?
The integral from 0 to 6r of sqrt(36r^2-y^2) is 9pir^2