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

The integral from 0 to 3 of sqrt((9-x^2)^3) is (243pi)/(16)

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

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

\frac{243 π}{16}

Decimal

47.71293 \dots

Solution steps

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

(9 - x^{2})^{3} = (9 - x^{2})^{\frac{3}{2}},

assuming

(9 - x^{2}) \geq 0

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

Apply Integration By Parts

: [x (9 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 9 - x^{2} d x]_{0}^{3}

= [x (9 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 9 - x^{2} d x]_{0}^{3}

\int - 3 x^{2} 9 - x^{2} d x = - \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))

= [x (9 - x^{2})^{\frac{3}{2}} - (- \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x))))]_{0}^{3}

Simplify

= [x (9 - x^{2})^{\frac{3}{2}} + \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))]_{0}^{3}

Compute the boundaries:

\frac{243 π}{16}

= \frac{243 π}{16}

\int_{- 1}^{1} (x^{2} - \frac{1}{3}) d x

\int_{- 3}^{- 1} (x^{3} + 6 x^{2} + 9 x + 9) d x

\int_{0}^{\frac{π}{4}} cos (2 y) d y

\int_{0}^{1} (2 + 3 x) d x

\int_{0}^{1} sin (u) d u

What is the integral from 0 to 3 of sqrt((9-x^2)^3) ?
The integral from 0 to 3 of sqrt((9-x^2)^3) is (243pi)/(16)