A pool shaped as a hemisphere with radius 5 m is filled to the top. How much work is required to remove the top 2 m of water (by pumping it out over the edge). The density of water is approximately 1000 kg/m 3 . Carefully explain how you obtain the integral
need help ASAPP
- Thread starter mini..45
- Start date
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
The work done lifting weight m a height h is mh. Imagine dividing the hemisphere into many thin layers, each of thickness "dy". The hemispheric pool can be modeled by \(\displaystyle x^2+ y^2= 25\) so that at height y, \(\displaystyle x= \sqrt{25- y^2}\) and x is the radius of the circular cross section. The area of that cross section is \(\displaystyle \pi r^2= \pi x^2= \pi(25- y^2)\). The volume of a circular "slab" of thickness dy is \(\displaystyle \pi(25- y^2)dy\) and the weight of that volume is \(\displaystyle 1000g\pi(25- y^2)\). That 'slab" has to be lifted a height y so the work done is \(\displaystyle 1000g\pi y(25- y^2)dy= 1000g\pi(25y- y^3)dy\) and the work done lifting the top 2 m of water is \(\displaystyle 1000g\pi\int_0^2 (25y- y^3)dy\).
(The density of water is given in "kg per cubic meter" which is mass, not weight. That is why I included "g".)
(The density of water is given in "kg per cubic meter" which is mass, not weight. That is why I included "g".)
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,215
work = [MATH]\int WALT[/MATH], where ...
[MATH]W[/MATH] is weight density
[MATH]A[/MATH] is the cross-sectional area of a representative horizontal "slice" of water
[MATH]L[/MATH] is the lift distance for a representative slice
[MATH]T[/MATH] is the slice thickness
with the center of the hemisphere's base at the origin, the work required to pump out the top 2 meters of water ...
[MATH]W = 1000g \cdot \pi \int_3^5 (25-y^2)(5 - y) \, dy [/MATH]
[MATH]W[/MATH] is weight density
[MATH]A[/MATH] is the cross-sectional area of a representative horizontal "slice" of water
[MATH]L[/MATH] is the lift distance for a representative slice
[MATH]T[/MATH] is the slice thickness
with the center of the hemisphere's base at the origin, the work required to pump out the top 2 meters of water ...
[MATH]W = 1000g \cdot \pi \int_3^5 (25-y^2)(5 - y) \, dy [/MATH]
The equation of the sphere is [MATH]x^2 + y^2 + z^2 = r^2[/MATH]
In our case the equation will be [MATH]x^2 + y^2 + z^2 = 5^2[/MATH]
We will take the bottom part of the sphere
[MATH]z = -\sqrt{25 - x^2 - y^2}[/MATH] (equation of the hemisphere)
work [MATH]=[/MATH] change in potential energy [MATH]= \Delta mgz = mgz_f - mgz_i[/MATH]
where [MATH]z_i[/math] and [math]z_f[/MATH] are initial and final heights respectively.
[MATH]m[/MATH] is the mass of the water inside the hemisphere, and [MATH]dm[/MATH] is a very small mass of that water.
then
[MATH]gz_f \ dm - gz_i \ dm[/MATH] is a very small work
Now, we just have to set the integral
work [MATH]= \int gz_f \ dm - \int gz_i \ dm[/MATH]
[MATH]dm = \rho \ dV[/MATH]
work [MATH]= \rho g\int z_f \ dV - \rho g\int z_i \ dV[/MATH]
[MATH]= \rho g\int \int \int z_f \ dz \ dy \ dx - \rho g\int \int \int z_i \ dz \ dy \ dx[/MATH]
setting up the limits
Remember that the first work will be done from the lowest point of the pool till the surface, and the second work will be done from the lowest point till [MATH]z = -2[/MATH]. By subtracting them from each other, we get the work done from [MATH]z = -2[/MATH] to [MATH]z = 0[/MATH] which is 2 meters at the top of the pool. And at [MATH]z = -2[/MATH], the radius of the circle on [MATH]x-y[/MATH] plane is [MATH]\sqrt{21}[/MATH].
so, the integral will be (without subscripts)
[MATH]= \rho g \ 2\int_{0}^{5} 2\int_{0}^{\sqrt{25 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{0} z \ dz \ dy \ dx - \rho g \ 2\int_{0}^{\sqrt{21}}2\int_{0}^{\sqrt{21 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{-2} z \ dz \ dy \ dx[/MATH]
[MATH]= 4\rho g \ \int_{0}^{5} \int_{0}^{\sqrt{25 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{0} z \ dz \ dy \ dx - 4\rho g \ \int_{0}^{\sqrt{21}}\int_{0}^{\sqrt{21 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{-2} z \ dz \ dy \ dx[/MATH]
Evaluating this integral with an online calculator will save you a lot of time, and also remember that the answer will be negative because we are calculating the bottom part of the sphere. Therefore, use the absolute value in the final result.
In our case the equation will be [MATH]x^2 + y^2 + z^2 = 5^2[/MATH]
We will take the bottom part of the sphere
[MATH]z = -\sqrt{25 - x^2 - y^2}[/MATH] (equation of the hemisphere)
work [MATH]=[/MATH] change in potential energy [MATH]= \Delta mgz = mgz_f - mgz_i[/MATH]
where [MATH]z_i[/math] and [math]z_f[/MATH] are initial and final heights respectively.
[MATH]m[/MATH] is the mass of the water inside the hemisphere, and [MATH]dm[/MATH] is a very small mass of that water.
then
[MATH]gz_f \ dm - gz_i \ dm[/MATH] is a very small work
Now, we just have to set the integral
work [MATH]= \int gz_f \ dm - \int gz_i \ dm[/MATH]
[MATH]dm = \rho \ dV[/MATH]
work [MATH]= \rho g\int z_f \ dV - \rho g\int z_i \ dV[/MATH]
[MATH]= \rho g\int \int \int z_f \ dz \ dy \ dx - \rho g\int \int \int z_i \ dz \ dy \ dx[/MATH]
setting up the limits
Remember that the first work will be done from the lowest point of the pool till the surface, and the second work will be done from the lowest point till [MATH]z = -2[/MATH]. By subtracting them from each other, we get the work done from [MATH]z = -2[/MATH] to [MATH]z = 0[/MATH] which is 2 meters at the top of the pool. And at [MATH]z = -2[/MATH], the radius of the circle on [MATH]x-y[/MATH] plane is [MATH]\sqrt{21}[/MATH].
so, the integral will be (without subscripts)
[MATH]= \rho g \ 2\int_{0}^{5} 2\int_{0}^{\sqrt{25 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{0} z \ dz \ dy \ dx - \rho g \ 2\int_{0}^{\sqrt{21}}2\int_{0}^{\sqrt{21 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{-2} z \ dz \ dy \ dx[/MATH]
[MATH]= 4\rho g \ \int_{0}^{5} \int_{0}^{\sqrt{25 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{0} z \ dz \ dy \ dx - 4\rho g \ \int_{0}^{\sqrt{21}}\int_{0}^{\sqrt{21 - x^2}} \int_{-\sqrt{25 - x^2 - y^2}}^{-2} z \ dz \ dy \ dx[/MATH]
Evaluating this integral with an online calculator will save you a lot of time, and also remember that the answer will be negative because we are calculating the bottom part of the sphere. Therefore, use the absolute value in the final result.