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

Answer:

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

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

Step-by-step 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

$dV=\pi(\sqrt{13^2-y^2})^2dy$

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

We need to calculate the mass of the strip

Using formula of mass

$W=\rho dVgy$

Put the value into the formula

$W=\pi({169-y^2})dy\times62.5\ y$

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

Using formula of work done

$W= \int_{0}^{11}(\pi({169-y^2})dy\times62.5\ y)$

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

(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

$dV=\pi(\sqrt{13^2-(11-y)^2})^2dy$

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

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

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

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