A tank, shaped like a cone has height 11 meter and base radius 4 meter. It is placed so that the circular part is upward. It is full of water, and we have to pump it all out by a pipe that is always leveled at the surface of the water. Assume that a cubic meter of water weighs 10 000 N

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Answer:

The tank is a cone, so the radius of the cross section increases linearly with h

At y=0, r=4 and at y=11, r=0

So

r(y)=4/11 (11−y)

A(y)=π[4(11−y)11]^2

Integral will be:

11∫0 (1000)(9.8)(4πy^2)(25)(11−y)dy

Step-by-step explanation: