How to Find the Area of a Truncated Ellipse?

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To find the area of the intersection between a cylinder and a plane forming a truncated ellipse, first determine the length of the perpendicular bisector to the chord using the Pythagorean theorem. Next, calculate the width of the circle based on the distance from the center and the height from the position in z. The area of the rectangle formed by the chords is 60√3, while the semi-ellipses have a height of 1 and a length of 6√3. The area of the two semi-ellipses can be calculated using the formula for the area of an ellipse, πab, where a and b are the semi-axes. An integral can be set up to express the width in terms of theta, leading to a more precise calculation of the area.
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Does anyone know how to find the area of an intersection between a cylinder of height 8 and radius 6 and a plane that passes through the cylinder, forming a chord of 10 units at the top and bottom faces of the cylinder? The area of intersection curves with the cylinder, forming a truncated ellipse, not a rectangle. I'm thinking you could divide the truncated ellipse into a rectangle and two sections of a circle, and find the rectangle by pretending the two chords are opposite edges of a rectangular prism, but I don't know how to find the width of the prism. Can anyone help?
 
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First find the length of the perpendicular bisector from the center to the chord using Pythagorean thm.
Then find the width of the circle based on distance from the center (r cos).
Then find the distance from center (height) based on position in z. (linear)
 
Ok so I figured out the area of the rectangle, 60 sqrt 3, and I know that the height of each semi-ellipse is 1, and their length is 6 sqrt 3, so how would I find the areas of the two semi-ellipses?
 
area of an ellipse is ##\pi a b ## where a is minor semi-axis and b is major semi-axis.
 
What if you recast width in terms of theta where theta goes from ## -sin^{-1}\frac{\sqrt 11}{6}## to ##sin^{-1}\frac{\sqrt 11}{6}##?
then you will have an integral
## \int_{-sin^{-1}\frac{\sqrt 11}{6}} ^{sin^{-1}\frac{\sqrt 11}{6}} width(\theta)length(\theta)d\theta##.
 

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