Finding the volume of liquid in an inclined cylinderA 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^{circ}) or 1.2164. I'm not sure how to figure out the changing lengths of the rectangles.Am I on the right track? ∫ 0 4.7 sin ⁡ ( 15 ° ) 6.4008 ? d y

Question

Finding the volume of liquid in an inclined cylinderA 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^{circ}) or 1.2164. I&#039;m not sure how to figure out the changing lengths of the rectangles.Am I on the right track?      ∫          0              4.7      sin      ⁡      (      15              °            )        6.4008  ?  d  y

Kasen Schroeder · Accepted Answer

Step 1Here is way how to get the volume of the empty part with a single integral. You will need- the area of an ellipse with semi-major and semi-minor axes a and b:  A  =  π  a  bThe water surface corresponds to a cut of a plane with the cylinder and is, hence, an ellipse. Now, the volume above water level can be modelled using ellipses. Just imagine that the angle between the plane and the top of the cylinder decreases.Let       ν    0   be the angle between the water surface and the top of the cylinder.Now, imagine for each   ν  ∈  [  0  ,      ν    0    ] a plane cuts the cylinder as described above. You sum up the areas of all the ellipses created by the cuts:      A          ν        =  π      a          ν            b          ν      If r is the radius of the cylinder, the main axes are      b          ν        =  r       and         a          ν        =      r          cos      ⁡              ν            Step 2Now, you get an infinitesimal volume slice (like an elliptic wedge) with respect to   ν as folllows:- the infinitesimal height at the end of the slice is             2      r              cos      ⁡      ν        d  ν (at the end of the major axis)- the vertical section of an elliptic wedge is an infinitesimal right triangle (similar to the calculation of area in polar coordinates), which gives rise to a factor of       1    2    d      V          ν        =      A          ν            1    2              2      r              cos      ⁡      ν        d  ν  =            π              r        3                            cos        2            ⁡      ν        d  νSo, the volume of the empty part is  V  =      ∫    0                  ν        0              d      V          ν        =  π      r    3        ∫    0                  ν        0                  1                  cos        2            ⁡      ν        d  ν  =  π      r    3    tan  ⁡      ν    0

Trystan Castaneda · Accepted Answer

Step 1If 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.Step 2Using this I get      ∫    0          6.4008        (  .20795  x  +  10.22291  )  (  2      10.24256    −    (    3.2004    −    x          )      2        )  d  x  =  350.3675  c      m    3

A right circular cylinder is at an incline of 15^circ 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?

Open question

Answer & Explanation

New Questions in High school geometry