Question

Find the area of the region bounded by the following functions: 𝑦 = 𝑥 + 3; 𝑦 = 3 − 2𝑥; 𝑦 = (𝑥 − 1)^2 − 1. The upper part of the ellipse (𝑥−4)^2/9 + (𝑦−3)^2/4 = 1; the lower part of the ellipse (𝑥−4)^2/9 + (𝑦−1)^2/4 = 1.

          Find the area of the region bounded by the following functions: 𝑦 = 𝑥 + 3; 𝑦 = 3 − 2𝑥; 𝑦 = (𝑥 − 1)^2 − 1. The upper part of the ellipse (𝑥−4)^2/9 + (𝑦−3)^2/4 = 1; the lower part of the ellipse (𝑥−4)^2/9 + (𝑦−1)^2/4 = 1.
        
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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the area of the region bounded by the following functions: 𝑦 = 𝑥 + 3; 𝑦 = 3 − 2𝑥; 𝑦 = (𝑥 − 1)^2 − 1. The upper part of the ellipse (𝑥−4)^2/9 + (𝑦−3)^2/4 = 1; the lower part of the ellipse (𝑥−4)^2/9 + (𝑦−1)^2/4 = 1.
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Transcript

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00:01 Okay, so if we have given a question, we have to find the area bounded by the curve, which is ellipse, which is given as x squared divided by 4 plus y squared divided by 9 is equal to 1.
00:16 So if you will compare this equation with a standard equation of an ellipse, which is x squared divided by a square plus y squared divided by b square is equal to 1.
00:26 If you will compare we'll get as plus minus 2 and b is plus minus 3 so absolute value of b is greater so this will be a semi major axis and which is along y axis means the parable uh sorry the ellipse is vertical so this is the ellipse and this will be your 3 this will be vertices we are writing which is minus 3 this is 2 and this will be minus 2 this is obviously 0.
01:05 So this elipse as you can see in the figure this is symmetrical about both the x is symmetrical about both the xes.
01:20 The area bounded by this curve is equal to the area coming in the first quadrant is equal to the area coming in the second quadrant which is equal to the third quadrant and the fourth quadrant means each areas are equal to another.
01:37 So if you can find the area of the curve bounded by coming in the first quadrant and we will multiply it by four to get the final answer.
01:49 So we will find the area.
01:51 Let's name it say o a, b, o, which is nothing but the integration of, to find the area using integration we have to take one strip...
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