Find the area of the ellipse

Question

Estimate the area of the ellipse given by the equation 4x^2+y^2=49 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, 3.5 ] into four equal subintervals and use the left endpoint of each subinterval.

 

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

Answer 1

An ellipse is in the form  X²/A² + Y²/B² = 1    

 

(for A>B)

where A is the semi-major axis   (1/2 width of ellipse)  

    and B is the semi-minor axis   (1/2 height of ellipse)

 

(for A<B)
where A is the semi-minor axis  (1/2 width of ellipse)
    and B is the semi-major axis  (1/2 height of ellipse)

 

so dividing both sides of your equation by 49 we get:

4X²/49 + Y²/49 = 1   OR    X²/(49/4) + Y²/49 = 1  OR  X2 / (7/2)² + Y2/7² = 1

 

THEREFORE:   A = 7/2 or 3.5   and B = 7

The area of an ellipse =  π·A·B   so substituting values for A and B we get π·A·B =  3.14159 · (3.5) ·(7) = 76.97

 

BUT, to approximate by their method we will use only the 1st quadrant and approximate the area, then multiply by 4. 

Multiply by 4, since the ellipse is centered at 0, and each quadrant has 1/4 of the ellipse

 

since A = 1/2 the width of the ellipse, and A above = 3.5, we divide 3.5 into 4 equal parts, each part = 7/8

subintervals will then be  1)  0-7/8   2) 7/8 - 14/8   3) 14/8 - 21/8  4) 21/8 - 28/8   (note: 28/8 = 3.5)

 

They say use the left end point of each interval, meaning use that as our X value.

 

So the X pts will be  0, 7/8, 14/8, 21/8.   Now we just need the Y pts

 

if we solve the original equation for Y we first put arrange it as    Y² = 49 - 4X²

So Y = √(49 - 4X²)

 

So our Y pts at our four  X pts  will be 7, 6.77, 6.06, 4.63     (example: at X = 7/8, Y = √(49 - 4(7/8)²)   OR Y = 6.77)

 

We now have four intervals, all 7/8 in width, and of heights 7, 6.77, 6.06, 4.63

 

Therefore, approx area of 1/4 of our ellipse (1st quadrant only) =  (7/8)·(7 + 6.77 + 6.06 + 4.63) = 21.4

 

So, approx 1/4 of our ellipse = 21.4,  

therefore the estimate for the whole ellipse we mult this number by 4 OR 21.4 · 4  = 85.6

So using the method summing 4 rectangles gives us the approx area = 85.6

 

Our estimate of 85.6 is a bit larger than the exact value 76.97 we found above, because we used the left most point of each interval, has the largest Y value in that interval. The actual ellipse at that point slopes downward from the top of the ellipse at X=0. The rectangle sub-interval therefore over estimates the area under the ellipse.

 

Our estimate would get closer to the exact value if we subdivided the interval into more subintervals.

 

 

Another method that would give a better estimate for the same number of intervals, would be sum polygons instead of rectangles. Then we just use the AVERAGE height of each sub-interval, instead of just using the leftmost point. 

 

So for our Y pts, 7, 6.77, 6.06, 4.63 we'd use instead  (7+6.77)/2,  (6.77+6.06)/2,  (6.06+4.63)/2, (4.63+0)/2  

(note: the last pt is 0 at X = 3.5)

 

so then we'd have Y pts of  6.88, 6.42, 5.35, 2.32 of the same width of 7/8

The estimate then of the 1/4 ellipse would be = 7/8 · (6.88+6.42+5.35+2.32) = 18.35

Multiplying by 4, for the whole ellipse would =  73.40 which is closer to the exact 76.97