What are the limits of integration, volume and distance?vertical shift down 2 units … what happens to the parabola function?
If the origin was at the top of the water, then I would move the origin down 2 units and be back at the original problem and solve that one. If you insist on not moving the origin, then you need to decide what the equation will be of the container, from which y values will each slab of water move from and to?What are the limits of integration, volume and distance?
Thank you for the indications. I am not planning to move it but I am having a lot of trouble setting up the equation and limits of integration. Could you please give some ideas to start?If the origin was at the top of the water, then I would move the origin down 2 units and be back at the original problem and solve that one. If you insist on not moving the origin, then you need to decide what the equation will be of the container, from which y values will each slab of water move from and to?
You need the EXACT same information that you needed to do the original problem.
Try to solve it, it will be fun. DRAW a diagram of the new setting!
I asked you a question … the parabola function with vertex at the origin is [imath]y=\dfrac{x^2}{2}[/imath].What are the limits of integration, volume and distance?
I did give you an idea on how to start. DRAW A DIAGRAM. In the original graph you were able to get the information that you needed to do the problem from the diagram. This same information will be in the new diagram.Thank you for the indications. I am not planning to move it but I am having a lot of trouble setting up the equation and limits of integration. Could you please give some ideas to start?