What is the integral from 0 to 4 of pi(sqrt(4-y))^2 ?

The integral from 0 to 4 of pi(sqrt(4-y))^2 is 8pi

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integral from 0 to 4 of pi(sqrt(4-y))^2

\int_{0}^{4} π (4 - y)^{2} d y

8 π

Decimal

25.13274 \dots

Solution steps

\int_{0}^{4} π (4 - y)^{2} d y

Take the constant out:

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

= π \cdot \int_{0}^{4} (4 - y)^{2} d y

Simplify

(4 - y)^{2} : 4 - y

= π \cdot \int_{0}^{4} 4 - y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π (\int_{0}^{4} 4 d y - \int_{0}^{4} y d y)

\int_{0}^{4} 4 d y = 16

\int_{0}^{4} y d y = 8

= π (16 - 8)

Simplify

= 8 π

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What is the integral from 0 to 4 of pi(sqrt(4-y))^2 ?
The integral from 0 to 4 of pi(sqrt(4-y))^2 is 8pi