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How do you find the volume of water in a hemispherical bowl of radius A and depth H using integration?
Wiki User
∙ 15y ago
Slice the bowl horizontally into circles, then integrate the area of the circles. The area of each circle is (pi * r^2). The height of each slice is dh. The 1st (bottom) circle is r=0. The r^2 of each circle-slice is (2*A*h-h^2), where A is the spherical radius, and h is the variable height of any given slice. At the top of the water level, (r^2=2*A*H-H^2). Integrate the area over the interval h=0->H as follows: V=pi * integral[(2*A*h - h^2) dh]; h=0->H to yield V=pi * (2*A*h^2 / 2 - h^3 / 3); h=0->H V=pi * (A*H^2 - H^3 / 3). As a check, plug the full diameter (2*A) in for H. If you did the integration correctly, you will get the full volume of the sphere, (4/3 * pi * A^3).
Wiki User
∙ 15y ago
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A hemi-spherical vessel has internal radius 0.5 m It is initially empty Water flows in at a constant rate of 1 liter per second Find an expression for the depth of the water after t seconds?
Since the vessel is hemispherical, its volume can be given by:V=((4/3)(pi)r3)/2V=(2/3)(pi)r3where r is the radius of the vessel.Since water is flowing into the vessel at a constant rate of 1 L/s, the volume of water in the vessel is thereby increasing at a constant rate of 1 L/s.By deriving the volume equation for the vessel with respect to time, we can equate the rate of change of the volume of water to the rate of change of the radius of the surface of water:dV/dt = (2/3)(pi)(3r2)(dr/dt)You must derive implicitly, so r3 derives down to 3r2(dr/dt) since the radius is also in itself a function of time. This equation can be cleaned up:dV/dt = 2(pi)r2(dr/dt)By solving for dr/dt, we get an expression for rate of change of the radius of the surface of water.dr/dt = (dV/dt)/(2(pi)r2)From the problem, we know that dV/dt is 1 L/s, and the radius of the hemisphere is a constant 0.5 m. We can substitute these known values into the equation:dr/dt = 1/(2(pi)(0.5)2)dr/dt = 2/piThis is the rate of change of the radius of the surface of water. The rate of change is a constant, which is important. Since it is constant, you can simply multiply this rate of change by a quantity of time to find the radius of the water level at any specific time. This is analogous to multiplying a constant velocity times a quantity of time to know an object's position at that time (a rate of change times an amount of change). We know that the vessel has an overall radius of 0.5 m, so the radius of the surface of water cannot exceed 0.5m.(dr/dt)= 2/pitherefore, depth at time t, D= 2t/piThis model gives the depth of the water (D) at any given time (t). As t increases, D(t) will return larger and larger values, which is expected since the water depth will increase as more water flows in.
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