What is the integral from 0 to 2 of r^2sqrt(16-r^2) ?

The integral from 0 to 2 of r^2sqrt(16-r^2) is 32(pi/6-(sqrt(3))/8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of r^2sqrt(16-r^2)

\int_{0}^{2} r^{2} 16 - r^{2} d r

32 (\frac{π}{6} - \frac{3}{8})

Decimal Notation

9.82695 \dots

Solution steps

\int_{0}^{2} r^{2} 16 - r^{2} d r

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 256 sin^{2} (u) cos^{2} (u) d u

= \int_{0}^{\frac{π}{6}} 256 sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= 256 \cdot \int_{0}^{\frac{π}{6}} sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 256 \cdot \int_{0}^{\frac{π}{6}} \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= 256 \cdot \frac{1}{8} \cdot \int_{0}^{\frac{π}{6}} 1 - cos (4 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

= 256 \cdot \frac{1}{8} (\int_{0}^{\frac{π}{6}} 1 d u - \int_{0}^{\frac{π}{6}} cos (4 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (4 u) d u = \frac{3}{8}

= 256 \cdot \frac{1}{8} (\frac{π}{6} - \frac{3}{8})

Simplify

256 \cdot \frac{1}{8} (\frac{π}{6} - \frac{3}{8}) : 32 (\frac{π}{6} - \frac{3}{8})

= 32 (\frac{π}{6} - \frac{3}{8})

\int_{- \infty}^{0} - x e^{- x^{2}} d x

\int_{1}^{\infty} e^{- 3 n} d n

\int_{0}^{π} x cos (d) x d x

\int_{- 3}^{3} x^{2} + 1 d x

\int_{- 2}^{2} x^{2} + 5 d x

What is the integral from 0 to 2 of r^2sqrt(16-r^2) ?
The integral from 0 to 2 of r^2sqrt(16-r^2) is 32(pi/6-(sqrt(3))/8)