Question

A tank in the shape of a hemisphere with a radius r = 13 ft is full of water to a depth of 11 ft. Set up the integral that would find the work W required to pump the water out of the spout. Use the location of the origin and the direction of the positive axis as specified in the different parts. hemisphere

Note: Use 62.5 lb/ft^3 as the weight of water.
a. The origin is at the top of the tank with positive y variables going down.
b. The origin is at the bottom of the tank with positive y-values going up.

Answer 1

(a). The integral is
`W=62.5\pi\int_(0)^(11)(169-y^2)ydy`

(b). The integral is
`W=62.5\pi\int_(0)^(11){48+22y-y^2(11-y)}dy`

Explanation:

Given that,

Radius = 13 ft

Depth = 11 ft

Weight of water = 62.5 lb/ft³

(a). The origin is at the top of the tank with positive y variables going down.

We need to calculate the volume of the strip

Using formula of volume

`dV=\pi r^2dy`

Put the value into the formula

`dV=\pi(√(13^2-y^2))^2dy`

`dV=\pi({169-y^2})dy`

We need to calculate the mass of the strip

Using formula of mass

`W=\rho dVgy`

Put the value into the formula

`W=\pi({169-y^2})dy*62.5\ y`

We need to calculate the work done to pump the water out of the spout

Using formula of work done

`W= \int_(0)^(11)(\pi({169-y^2})dy*62.5\ y)`

`W=62.5\pi\int_(0)^(11)(169-y^2)ydy`

(b). The origin is at the bottom of the tank with positive y-values going up

We need to calculate the volume of the strip

Using formula of volume

`dV=\pi r^2dy`

Put the value into the formula

`dV=\pi(√(13^2-(11-y)^2))^2dy`

`dV=\pi(169-(121-22y+y^2))dy`

`dV=\pi(169-121+22y-y^2)dy`

`dV=\pi(48+22y-y^2)dy`

We need to calculate the mass of the strip

Using formula of mass

`W=\rho Vg`

Put the value into the formula

`W=dV\rho g(11-y)`

`W=\pi(48+22y-y^2)dy\rho g(11-y)`

`W=62.5\pi(48+22y-y^2)(11-y)dy`

We need to calculate the work done to pump the water out of the spout

Using formula of work done

`W=\int_(0)^(11){62.5\pi(48+22y-y^2)(11-y)dy}`

`W=62.5\pi\int_(0)^(11){48+22y-y^2(11-y)}dy`

Hence, (a). The integral is
`W=62.5\pi\int_(0)^(11)(169-y^2)ydy`

(b). The integral is
`W=62.5\pi\int_(0)^(11){48+22y-y^2(11-y)}dy`