+129657 We can split this into two integrals
0 0
∫ 4 dx + ∫ √ [49 - x^2 ] dx
-7 -7
0
The first is easy.....we have 4x ] = 0 - (-28) = 28
-7
For the second part we have
0
∫ √ [49 - x^2 ] dx
-7
Let a = 7 → a^2 = 49
Let x = a cos θ → 7 cos θ → x^2 = 49cos^2 θ and dx = -7 sin θ dθ
Now.....when x = 0 , 0 = -7cos θ .... so θ = pi/2
And when x = -7 , - 7 = 7cos θ .→ -1 = cos θ ....so θ = pi
Making the following substitutions, we have
pi/2
∫ √ [49 - 49cos^2 θ ] * -7 sin θ dθ
pi
pi/2
∫ 7 √ [1 - cos^2 θ ] * -7 sin θ dθ
pi
pi/2
-49 ∫ √ [1 - cos^2 θ ] * sin θ dθ
pi
pi/2
-49 ∫ sin θ * sin θ dθ
po
pi/2
-49 ∫ sin^2 θ dθ note ... cos2θ = 1 - 2sin^2θ → sin^2 θ = [ 1 - cos2 θ ] / 2 = 1/2 - [cos2 θ] / 2
pi
So we have
pi/2 pi/2
-49 [ ∫ 1/2 dθ - ∫ [cos 2 θ] / 2 dθ ] =
pi pi
-49 [ (1/2) [ pi/2 - pi) - 1/4 [ sin ( 2*pi/2) - sin (2pi) ] =
-49 [ (1/2)(-pi/2) - (1/4)*[0 - 0] =
[-49] [ -pi/4 ] ≈ 38.4845
Putting tis all together, we have 28 + 38.4845 ≈ 66.4845
Note....that we didn't really need to use Calculus on the second half of the problem....this is just the area of a quarter circle with a radius of 7 = (49/4)*pi ≈ 38.4845.......
