What is the integral from 0 to (2pi)/4 of integral from 1 to 2 of sqrt(4-r^2)*r ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to (2pi)/4 of integral from 1 to 2 of sqrt(4-r^2)*r

\int_{0}^{\frac{2 π}{4}} \int_{1}^{2} 4 - r^{2} \cdot r d r d θ

\frac{3 π}{2}

Decimal

2.72069 \dots

Solution steps

\int_{0}^{\frac{2 π}{4}} \int_{1}^{2} 4 - r^{2} r d r d θ

\int_{1}^{2} 4 - r^{2} r d r = 3

= \int_{0}^{\frac{2 π}{4}} 3 d θ

\int_{0}^{\frac{2 π}{4}} 3 d θ = \frac{3 π}{2}

= \frac{3 π}{2}

What is the integral from 0 to (2pi)/4 of integral from 1 to 2 of sqrt(4-r^2)*r ?
The integral from 0 to (2pi)/4 of integral from 1 to 2 of sqrt(4-r^2)*r is (sqrt(3)pi)/2