You’re welcome! Glad the video could help. I have lots of other Calculus 2 videos in a playlist that you can check out! And I will definitely be making more videos for more courses in the future 👍
I am confused. In earlier example, you calculated the area of the object that is submerged in water. In the later example, you calculate the area of water in the tank but not the area of the tank.
If I understand your question/confusion correctly, here is what is important to remember: with these types of problems we are looking at the fluid pressure/force on a two-dimensional surface or face of an object, not necessarily an entire three dimensional object. In the first couple examples we look at a flat object submerged in water and are interested in the fluid force on the object's face. We are still doing this same exact thing in the last example, except the "submerged object" is just a vertical side of a tank of water. Since the tank is filled with water, a vertical side of the tank will experience fluid pressure/force on the side inside the tank. So we are still dealing with a "submerged" object, if that makes sense. Just remember that we are only looking at one side of the tank, not all sides of the tank, or the entire tank itself. Just one vertical side. I know the scenarios are slightly different, but the same idea is still at play. Let me know if this helps or if you need any further explanation on this topic!
In the 11.18 vertical surface example, you treat the plate as a two dimensional object and calculated the area of the plane. But in the tank example, you treat the side of tank as one dimensional thing and calculate the area of water at each depth instead. If the object is a dam, will you approach the problem as if it is like the vertical surface or like the tank
@@haileykk1035 So the difference between my approach in the two examples is based on the shape of the surfaces/objects we are working with. For the example with the plane, the area is as simple as length times height since the shape is that of a rectangle (going back to the demonstration at around the 11 minute mark, remember that we are looking at a representative rectangle for these shapes). The length in that scenario is constant (as it also is in Example 2) However the length of the side of the tank in Example 3 is NOT constant for the representative rectangle since the side of the tank is a trapezoidal shape, not a rectangle. Both objects are still being treated as 2D objects, it's just that the shape of those 2D objects are different, so we have to take a different approach. The way you calculate area of a trapezoid is not the same way that you calculate the area of a rectangle, and so we have to reflect that in our integral. So in the case of a dam being your object, your approach is going to depend on the shape of the dam. Is it a rectangle? A trapezoid? A triangle? etc. Does that help?
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thank you, the explanation is super easy and understandable please continue making more videos like this
You’re welcome! Glad the video could help. I have lots of other Calculus 2 videos in a playlist that you can check out! And I will definitely be making more videos for more courses in the future 👍
Thanks for the explanation sir
You're welcome!
I am confused. In earlier example, you calculated the area of the object that is submerged in water. In the later example, you calculate the area of water in the tank but not the area of the tank.
If I understand your question/confusion correctly, here is what is important to remember: with these types of problems we are looking at the fluid pressure/force on a two-dimensional surface or face of an object, not necessarily an entire three dimensional object. In the first couple examples we look at a flat object submerged in water and are interested in the fluid force on the object's face. We are still doing this same exact thing in the last example, except the "submerged object" is just a vertical side of a tank of water. Since the tank is filled with water, a vertical side of the tank will experience fluid pressure/force on the side inside the tank. So we are still dealing with a "submerged" object, if that makes sense. Just remember that we are only looking at one side of the tank, not all sides of the tank, or the entire tank itself. Just one vertical side. I know the scenarios are slightly different, but the same idea is still at play. Let me know if this helps or if you need any further explanation on this topic!
In the 11.18 vertical surface example, you treat the plate as a two dimensional object and calculated the area of the plane. But in the tank example, you treat the side of tank as one dimensional thing and calculate the area of water at each depth instead. If the object is a dam, will you approach the problem as if it is like the vertical surface or like the tank
@@haileykk1035 So the difference between my approach in the two examples is based on the shape of the surfaces/objects we are working with. For the example with the plane, the area is as simple as length times height since the shape is that of a rectangle (going back to the demonstration at around the 11 minute mark, remember that we are looking at a representative rectangle for these shapes). The length in that scenario is constant (as it also is in Example 2) However the length of the side of the tank in Example 3 is NOT constant for the representative rectangle since the side of the tank is a trapezoidal shape, not a rectangle. Both objects are still being treated as 2D objects, it's just that the shape of those 2D objects are different, so we have to take a different approach. The way you calculate area of a trapezoid is not the same way that you calculate the area of a rectangle, and so we have to reflect that in our integral. So in the case of a dam being your object, your approach is going to depend on the shape of the dam. Is it a rectangle? A trapezoid? A triangle? etc. Does that help?