What is the integral from 0 to 6 of sqrt(9-(x-3)^2) ?

The integral from 0 to 6 of sqrt(9-(x-3)^2) is (9pi)/2

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integral from 0 to 6 of sqrt(9-(x-3)^2)

\int_{0}^{6} 9 - (x - 3)^{2} d x

\frac{9 π}{2}

Decimal Notation

14.13716 \dots

Solution steps

\int_{0}^{6} 9 - (x - 3)^{2} d x

Apply u-substitution

: \int_{- 3}^{3} 9 - u^{2} d u

= \int_{- 3}^{3} 9 - u^{2} d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 9 cos^{2} (v) d v

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 9 cos^{2} (v) d v

Take the constant out:

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

= 9 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (v) d v

Rewrite using trig identities

= 9 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 v) d v

Apply the Sum Rule:

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

= 9 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v = π

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

= 9 \cdot \frac{1}{2} (π + 0)

Simplify

9 \cdot \frac{1}{2} (π + 0) : \frac{9 π}{2}

= \frac{9 π}{2}

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What is the integral from 0 to 6 of sqrt(9-(x-3)^2) ?
The integral from 0 to 6 of sqrt(9-(x-3)^2) is (9pi)/2