Why is the upper limit 4 and not 16?

Question

Find the area enclosed by f(y)=y/4 g(y)=y^1/2

My teacher wrote the integral from 0 to 4 why wouldn't it be 16 set them equal to each other and 4=4

Answer 1

area = the integral of (4x -x^2) = 2x^2 -x^3/3 evaluated from 0 to the x coordinate where the line and curve intersect at (4, 16). they intersect twice at (0,0) and at (4,16)

2(4^2) - (4^3)/3 = 32 - 64/3 = (96-64)/3 = 32/3 = 10 2/3 square units

or

you could set up the problem in terms of y and the integral of f(y)dy then you get limits of 0 and 16

area = integral of (sqr(y) - y/4)dy = 2y^1.5)/3 - (y^2)/8 evaluated from 0 to 16

= (2/3)16^1.5 - (16^2)/8 = (2/3)64 - 32 = 128/3 - 96/3 = (128-96)/3 = 32/3 = 10 2/3 square units

same answer either way

Answer 2

The first equation is:

x = y/4

or y = 4x

The second equation is:

x = √y

or y = x², x ≥ 0

The intersection is (4,16)

The area in terms of x integral is:

∫ (4x - x²) dx, with lower limit 0 and upper limit 4

However, the area in terms of y integral is:

∫ (√y - y/4) dy, with lower limit 0 and upper limit 16

So, your professor is correct if he has set-up the integral in terms of x. Both integrals will give you the same answer.