Answer:
(a). The integral is [tex]W=62.5\pi\int_{0}^{11}(169-y^2)ydy[/tex]
(b). The integral is [tex]W=62.5\pi\int_{0}^{11}{48+22y-y^2(11-y)}dy[/tex]
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
[tex]dV=\pi r^2dy[/tex]
Put the value into the formula
[tex]dV=\pi(\sqrt{13^2-y^2})^2dy[/tex]
[tex]dV=\pi({169-y^2})dy[/tex]
We need to calculate the mass of the strip
Using formula of mass
[tex]W=\rho dVgy[/tex]
Put the value into the formula
[tex]W=\pi({169-y^2})dy\times62.5\ y[/tex]
We need to calculate the work done to pump the water out of the spout
Using formula of work done
[tex]W= \int_{0}^{11}(\pi({169-y^2})dy\times62.5\ y)[/tex]
[tex]W=62.5\pi\int_{0}^{11}(169-y^2)ydy[/tex]
(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
[tex]dV=\pi r^2dy[/tex]
Put the value into the formula
[tex]dV=\pi(\sqrt{13^2-(11-y)^2})^2dy[/tex]
[tex]dV=\pi(169-(121-22y+y^2))dy[/tex]
[tex]dV=\pi(169-121+22y-y^2)dy[/tex]
[tex]dV=\pi(48+22y-y^2)dy[/tex]
We need to calculate the mass of the strip
Using formula of mass
[tex]W=\rho Vg[/tex]
Put the value into the formula
[tex]W=dV\rho g(11-y)[/tex]
[tex]W=\pi(48+22y-y^2)dy\rho g(11-y)[/tex]
[tex]W=62.5\pi(48+22y-y^2)(11-y)dy[/tex]
We need to calculate the work done to pump the water out of the spout
Using formula of work done
[tex]W=\int_{0}^{11}{62.5\pi(48+22y-y^2)(11-y)dy}[/tex]
[tex]W=62.5\pi\int_{0}^{11}{48+22y-y^2(11-y)}dy[/tex]
Hence, (a). The integral is [tex]W=62.5\pi\int_{0}^{11}(169-y^2)ydy[/tex]
(b). The integral is [tex]W=62.5\pi\int_{0}^{11}{48+22y-y^2(11-y)}dy[/tex]
