Integration stuff

Question

Set up a single integral with respect to y that would find the area bounded by y = 3 sqrt(4-x), y = 1/3x+6 and y = 0. I can evaluate it... just, set up the integral and find the bounds?

Answer 1

The curves y = (1/3)x + 6 and y = 3√(4-x) intersect at the point (0,6).

If y = (1/3)x + 6, then x = 3y - 18.

If y = 3√(4-x) , then y² = 9(4-x) = 36 - 9x. So, x = (36-y²)/9 = 4 - (1/9)y²

Area = ∫_{(0 to 6)} [(4 - (1/9)y²) - (3y - 18)]dy = ∫_{(0 to 6)} [22 - 3y - (1/9)y²]dy