Find the area of the

Question

K={(x,y)∈R/x^2+y^2<=1, o<=y<=x^2, x>=0}

Answer 1

The region to find the area of is in the figure to the right:

First, we must find the x value of the intersection point of y = x² and x² + y² = 1, and

given that x >= 0, we have the following: x² + (x²)² = 1, so (x²)² + x² - 1 = 0, so, by the quadratic formula, x² = ( -b ± √[ b² - 4*a*c ] ) / (2*1) = ( -1 ± √[ 1² - 4*1*(-1) ] ) / (2*1),

so x² = ( -1 ± √[1+4] ) / 2 = [ -1 ± √(5) ] / 2, but we must discard the negative value,

so that x is real, so since x² = [ -1 + √(5) ] / 2, then x = √[√(5) - 1] / √(2), but for

simplicity in our integrals, let's let b = √[√(5) - 1] / √(2), and then plug it in at the

end. Solving for y in the 2nd equation gives us y² = 1 - x², so then y = √(1 - x²).

Therefore, now we have that the area of that region is the following:

A = ∫ x² dx from 0 to b + ∫ √(1 - x²) dx from b to 1

For the 2nd integral below, let x = sin(t), so dx = cos(t) dt; when x = b,

t = arcsin(b); and when x = 1, t = arcsin(1) = pi/2, so now we have:

A = x³ / 3 from 0 to b + ∫ √[ 1 - sin²(t) ] cos(t) dt from arcsin(b) to pi/2

Now, for the 2nd integral below, we can use the trig identity

cos(t) = √[ 1 - sin²(t) ], and get the following:

A = (b³ / 3 - 0³ / 3) + ∫ cos²(t) dt from arcsin(b) to pi/2

Now, for the 2nd integral below, we can use the trig identity

cos²(t) = [ 1 + cos(2t) ] / 2, and get the following:

A = b³ / 3 - 0 + ∫ [ 1 + cos(2t) ] / 2 dt from arcsin(b) to pi/2

A = b³ / 3 + [ t + sin(2t)/2 ] / 2 from arcsin(b) to pi/2

Now, for the 2nd integral below, we can use the trig identity

sin(2t) = 2*sin(t)*cos(t) = 2*sin(t)*√[ 1 - sin²(t) ], and get the following:

A = b³ / 3 + { t + 2 * sin(t) * √[ 1 - sin²(t) ] / 2 ] } / 2 from arcsin(b) to pi/2

A = b³ / 3 + { pi/2 + sin(pi/2) * √[ 1 - sin²(pi/2) ] } / 2 -

{ arcsin(b) + sin(arcsin(b)) * √[ 1 - sin²(arcsin(b)) ] } / 2

A = b³/3 + [ pi/2 + 1 * √(1-1²) ] / 2 - [ √(b) + b * √(1-b²) ] / 2

A = b³/3 + pi/4 + √(1-1)/2 - arcsin(b)/2 - b*√([2/2] - [√(5)-1]/2) / 2

A = b³/3 + pi/4 + √(0)/2 - arcsin(b)/2 - b*√( [2-√(5)+1] / 2) / 2

A = b³/3 + pi/4 + 0 - arcsin(b)/2 - √([3-√(5)]/2) * b/2

and finally:

A = b³/3 + pi/4 - arcsin(b)/2 - √( [3-√(5)] / 2) * b/2, where b = √[ √(5) - 1 ] / √(2).

Answer 2

Hi Gabriela,

The video is a bit long, but this was quite the problem to tackle. I did notice that I made a small mistake. In the end, I forgot to carry the negative, so it should be a positive phi and not negative. Hope this helps. I would normally take much more time, but I couldn't make the video even longer. Good luck!