Find the average value?

Question

Find the average value of y=sqrt(64-x^2) on its domain.

 

Answer: 2pi

 

I know that the limits of integration is from -8 to 8 but how do I take the integral of sqrt(64-x^2) dx?

 

Please show all your work. 

Answer 1

Find the average value of f(x) = √(64-x^2) on its domain.

I = integral{x1,x2}( f(x) dx)

The average value of f(x) is the height, h, of a rectangle with base x2-x1 whose area is I.

h = I/(x2-x1)

domain of f(x) is:
64-x^2 >= 0
64 >= x^2
8 >= |x|
-8 <= x <= 8

From precalculus we know that f(x) is the top semicircle of the circle x^2 + y^2 = 8^2. So its area is 1/2 pi 8^2 = 32 pi.

The average value of f(x) is then 32 pi/(8 – –8) = 2 pi.

We can also use a trig substitution to do the problem with calculus.

sin(θ) = x/8, cos(θ) = √(64-x^2)/8
x = 8 sin(θ), √(64-x^2) = 8 cos(θ)
dx = 8 cos(θ) dθ

-8 <= x <= 8 ==> -pi/2 <= θ <= pi/2

I = integral{-8,8}( √(64-x^2) dx)

I = integral{-pi/2,pi/2}( 8 cos(θ) 8 cos(θ) dθ)

I = 64 integral{-pi/2,pi/2}( cos^2(θ) dθ)

2 cos^2(θ) - 1 = cos(2θ)
cos^2(θ) = (1 + cos(2θ))/2

I = 64 integral{-pi/2,pi/2}( (1 + cos(2θ))/2 dθ)

I = 32 integral{-pi/2,pi/2}( (1 + cos(2θ)) dθ)

I = 32 [θ + sin(2θ)/2]{-pi/2,pi/2}

I = 32 [pi/2 + sin(pi)/2 - (-pi/2) - sin(-pi)/2]

I = 32 [pi/2 + pi/2] = 32 pi

The average value of f(x) is then 32 pi/(8 – –8) = 2 pi.

Answer 2

 

 

 

integral of sqrt(a^2-x^2) dx =  1/2x (sqrt (a^2 - x^2)) + 1/2 a^2 * arctan (x / sqrt (a^2 - x^2))

 

The integral of sqrt(64-x^2) dx is the same as sqrt(8^2-x^2) dx

 

so integral of  sqrt(8^2-x^2) dx = 1/2x (sqrt (8^2 - x^2)) + 1/2 8^2 * arctan (x / sqrt (8^2 - x^2))

 

 

1/2x (sqrt (64 - x^2)) + 32 * arctan (x / sqrt (64 - x^2))

 

solve the rest