Question 1183646
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The desired area in each case illustrated below is the area of the quarter ellipse minus the area of the shaded triangle.  So, the larger the triangle, the  smaller the desired area, and the minimum triangle gives the maximum desired area and vice versa.  The area of the quarter ellipse without any subtractions is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int_0^3\,\frac{\sqrt{36\,-\,4x^2}}{3}\,dx]


You can use a simple trig substitution to solve this integral.  Hint: Take out the constant term and then let *[tex \Large x\ =\ 3\sin(u)] and *[tex \Large dx\ =\ 3\cos(u)du]

Then subtract the triangle areas illustrated:


*[illustration Intercept_2]


*[illustration Intercept_1.5]


*[illustration Intercept_1]


*[illustration Intercept_0.5]



																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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