Estimate the area of the ellipse given by the equation 4x2+y2=36 as follows: The

Question

Estimate the area of the ellipse given by the equation 4x2+y2=36 as follows:

The area of the ellipse is 4 times the area of the part of the ellipse in the first quadrant ( x and y positive).

Estimate the area of the ellipse in the first quadrant by solving for y in terms of x. Estimate the area under the graph of y by dividing the interval [0,3] into four equal subintervals and using the left endpoint of each subinterval. Be sure you draw a picture.

Don't forget to multiply your estimate for the area of the part of the ellipse in the first quadrant by 4 to get the entire area.

The area of the ellipse (using the above method) is approximately ?

Explanation / Answer

For the first quadrant area approximation: Solving for y yields f(x) = y = sqrt(25 - 4x^2). ?x = (2.5 - 0)/4 = 2.5/4 = 25/40 = 5/8. So, the interval is subdivided into {0, 5/8, 10/8, 15/8, 20/8 = 2.5}. Using left endpoints yields A ˜ ?x [f(0) + f(5/8) + f(10/8) + f(15/8)] = (5/8) [5 + sqrt(375/16) + sqrt(75/4) + sqrt(175/16)] ˜ 10.924. So, the area of the ellipse is approximately 4(10.924) ˜ 43.696. (For the record, the exact area is p * (5/2) * 5 ˜ 39.270.

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