A right circular cylinder is at an incline of 15° from the horizontal and the liquid is level with the lowest point of the top rim of the can. The radius is 3.2004 cm and the height is 11.938 cm. What is the volume of the liquid?
I believe I should use integration with cross-sections of rectangles. The width of each rectangle would be the diameter of the cylinder. I believe my limits of integration would be from 0 to 4.7$\cdot\sin(15°)$ or 1.2164. I'm not sure how to figure out the changing lengths of the rectangles.
Am I on the right track?
$$\int_{0}^{4.7\sin(15°)}6.4008?dy$$
Best Answer
If you rotate the cylinder so the base is on the x axis and the liquid level is at 15 deg (positive slope), then the height of the rectangles is given by the height of the cylinder minus the evaluation of the 15 deg slope of the level surface (linear equation). Limits of integration can be x = 0, to x = d (cylinder diameter).The width of the rectangle will be the chord length of a line cutting the circumference of the circular cross section as it moves from left to right.
Using this I get$$\int_0^{6.4008} (.20795x+10.22291)(2\sqrt{10.24256-(3.2004-x)^2})dx$$ $= 350.3675 cm^3$