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e^(i Pi)+1=0
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I'm an engineering student, and I'm making a formula for the volume of liquid in a cylinder that is cut in half diagonally. r is the radius, h is the height from the bottom to the top of the sloped flat bottom, and L is the height of the water within the cylinder. So when L = h you would expect the formula to be equal to h∏r^(2)/2.
z bar is the plane defining the bottom where the cylinder is "cut." It was derived using the points (-r,0,0), (r,0,h) and (0,r,h/2) so it passes through the xy plane at x = -r and touches h at x = r.
y bar is the standard equation for half a circle with radius r
a is the x value of the rightmost edge of the liquid as the tank is filled.
[itex]v = 2\int_{-r}^{a}\int_{0}^{\bar{y}}\int_{\bar{z}}^{h}dzdydx[/itex]
[itex]\bar{z} = \frac{hx}{2r}+\frac{h}{2}[/itex]
[itex]\bar{y} = \sqrt{r^{2}-x^{2}}[/itex]
[itex]a = \frac{2rL}{h}-r[/itex]
Solving leads to the following formula:
[itex]v = \frac{r^{2}h*arcsin(a/r)}{2}+\frac{ha\sqrt{r^{2}-a^{2}}}{2}-\frac{3\pi r^{2}h}{4}+\frac{h(r^{2}-a^{2})^{3/2}}{3r}[/itex]
[itex]a = \frac{2rL}{h}-r[/itex]
It's much simpler the leave a as a separate equation, solve for a and then enter it into the larger equation. If this formula were correct, you would expect it to equal zero when L equals zero, which it does. You would also expect it to equal h∏r^(2)/2 when L = h except it equals the negative of that. I've worked hard on this and am trying to get it right. Is something wrong with my setup? Thanks.
z bar is the plane defining the bottom where the cylinder is "cut." It was derived using the points (-r,0,0), (r,0,h) and (0,r,h/2) so it passes through the xy plane at x = -r and touches h at x = r.
y bar is the standard equation for half a circle with radius r
a is the x value of the rightmost edge of the liquid as the tank is filled.
[itex]v = 2\int_{-r}^{a}\int_{0}^{\bar{y}}\int_{\bar{z}}^{h}dzdydx[/itex]
[itex]\bar{z} = \frac{hx}{2r}+\frac{h}{2}[/itex]
[itex]\bar{y} = \sqrt{r^{2}-x^{2}}[/itex]
[itex]a = \frac{2rL}{h}-r[/itex]
Solving leads to the following formula:
[itex]v = \frac{r^{2}h*arcsin(a/r)}{2}+\frac{ha\sqrt{r^{2}-a^{2}}}{2}-\frac{3\pi r^{2}h}{4}+\frac{h(r^{2}-a^{2})^{3/2}}{3r}[/itex]
[itex]a = \frac{2rL}{h}-r[/itex]
It's much simpler the leave a as a separate equation, solve for a and then enter it into the larger equation. If this formula were correct, you would expect it to equal zero when L equals zero, which it does. You would also expect it to equal h∏r^(2)/2 when L = h except it equals the negative of that. I've worked hard on this and am trying to get it right. Is something wrong with my setup? Thanks.
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