I know you could solve when the fill is less than half full by using the same method for the empty space then subtracting the empty space from the max volume. But is there a way to combine the two ways into one formula that can be used for both applications?
it is convenient to put the center of the tank at the origin (0,0). But the height of the water is 7 feet total. So 4 feet below and 3 feet above because measuring the height of the water starts at the bottom of the tank
yes - good question, that's the next step - it's really the same type of approach - you'll just need to get the equation of the ellipse, with center of ellipse at center of tank, solve for x or y, and find an area in one quadrant, from the axis to boundary of ellipse, and then everything follows in the same way as with the circle.
For the filled tank with depth given by h, i think u had forgotten to include the area of the bottom half circle filled in calculating the area cross section filled
For that calculation, the integration begins at h = -4 which is at the very bottom of the tank. It's a good question because in the previous example I start the integration at h = 0. And then add the bottom half of the tank by computing area of the half circle. I integrated from h = -4, for the arbitrary depth h because the function p(h) that gives us percentage of the tank filled (as function of h) can be expressed more simply as a single integral, without that extra term for the area of the half circle.
What if we have calculated the volume of the empty tank then we multiplied the volume by the 0.7 which is 70% of the tank give us the tank filled volime?
no, it wouldn't really give you the volume of the fluid. That's because what you suggest ignores the cylindrical shape of the tank, and the fact that the cross sections are circular. This means, as the height of the fluid increases, a greater percentage of the cross section is filled.
It is helpful for Cal 2!!!!
is there a way to find the volume when the filling height is variable and can be less or more than half-filled?
I know you could solve when the fill is less than half full by using the same method for the empty space then subtracting the empty space from the max volume. But is there a way to combine the two ways into one formula that can be used for both applications?
How did you come to the conclusion of 3 for the upper limit?? What math was done? can you explain more?
it is convenient to put the center of the tank at the origin (0,0). But the height of the water is 7 feet total. So 4 feet below and 3 feet above because measuring the height of the water starts at the bottom of the tank
How did you get the equation of the curve? Is it just bc the radius is 4 and that’s 4 squared?
yes, that's right, equation of a circle, centered at origin, with radius 4
Thank you!!
how do you set this problem up for the tank being an ellipse?
yes - good question, that's the next step - it's really the same type of approach - you'll just need to get the equation of the ellipse, with center of ellipse at center of tank, solve for x or y, and find an area in one quadrant, from the axis to boundary of ellipse, and then everything follows in the same way as with the circle.
For the filled tank with depth given by h, i think u had forgotten to include the area of the bottom half circle filled in calculating the area cross section filled
For that calculation, the integration begins at h = -4 which is at the very bottom of the tank. It's a good question because in the previous example I start the integration at h = 0. And then add the bottom half of the tank by computing area of the half circle. I integrated from h = -4, for the arbitrary depth h because the function p(h) that gives us percentage of the tank filled (as function of h) can be expressed more simply as a single integral, without that extra term for the area of the half circle.
What if we have calculated the volume of the empty tank then we multiplied the volume by the 0.7 which is 70% of the tank give us the tank filled volime?
no, it wouldn't really give you the volume of the fluid. That's because what you suggest ignores the cylindrical shape of the tank, and the fact that the cross sections are circular. This means, as the height of the fluid increases, a greater percentage of the cross section is filled.
Fun! Ty!