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

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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 Notation

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}

\int \int x^{2} y^{2} d x d y

\int \int sin (x - y) d x d y

\int \int tan (x) d x d x

\int \int r d r d θ

\int \int \int z^{2} d x d y d z

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