What is the integral from 0 to 3 of 9*(9-y^2)^{1/2} ?

The integral from 0 to 3 of 9*(9-y^2)^{1/2} is (81pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of 9*(9-y^2)^{1/2}

\int_{0}^{3} 9 \cdot (9 - y^{2})^{\frac{1}{2}} d y

\frac{81 π}{4}

Decimal Notation

63.61725 \dots

Solution steps

\int_{0}^{3} 9 (9 - y^{2})^{\frac{1}{2}} d y

Take the constant out:

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

= 9 \cdot \int_{0}^{3} (9 - y^{2})^{\frac{1}{2}} d y

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 9 cos^{2} (u) d u

= 9 \cdot \int_{0}^{\frac{π}{2}} 9 cos^{2} (u) d u

Take the constant out:

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

= 9 \cdot 9 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 9 \cdot 9 \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

= 9 \cdot 9 \cdot \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

= 9 \cdot 9 \cdot \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

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

Simplify

9 \cdot 9 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{81 π}{4}

= \frac{81 π}{4}

\int_{0}^{x^{2} + y^{2}} 1 d z

\int_{4}^{5} 3 x x^{2} - 16 d x

\int_{2}^{3} \frac{7}{1 + x ^{2}} d x

\int_{- 5}^{0} (3 + ∣ x + 4 ∣) d x

\int_{- 1}^{3} 1 + (2 y + 1)^{2} d y

What is the integral from 0 to 3 of 9*(9-y^2)^{1/2} ?
The integral from 0 to 3 of 9*(9-y^2)^{1/2} is (81pi)/4