Solve for the following

Question

In the following exercises, find the area of the region by integrating (a) with respect to x and (b) with respect to y. (c) Compare your results. Which method is simpler? (d) In general, will this method always be simpler than the other one? Why or why not?

1. x=4-y²

x=y-2

2. y=x^2

y= 6-x

Answer 1

I'll do the first one.

1) you need to look at the graph of the two functions. The first is a parabola opening to the left (It is not a function in x because it is multivalued in y - this makes integrating y(x)dx harder than x(y)dy.)

The two points of intersection (the limits of the integration of (x₁ - x₂) dy) will occur when

4 - y² = y -2 or y² + y - 6 = 0 (y+3)(y-2) = 0 (x,y) = (0,2) and (-5,-3)

The integral for the area is the integral from y = -3 to 2 of ((4-y²) - (y-2))dy = int of -y² - y + 6 =

-y³/3 - y²/2 + 6y from -3 to 2 = (-8 - 27)/3 + (-4 + 9)/2 + 6(2+3) = 125/6

If you want to integrate in x

1) Rewrite the equations as y(x): y₁ = -sqrt(4-x) (because the area is bounded by the negative branch of the parabola) and y₂ = x + 2 (with y₂ above y₁

2) The area integral must be split in two because the difference in functions only holds from x = -5 to 0. There is another quirk: the difference is y₂ - (-sqrt(4-x)). From 0 to the apex of the parabola (at (4,0)) the integral is double the area to the y₁ curve.

Int from -5 to 0 (sqrt(4-x) + (2+x))dx + 2*Int from 0 to 4 of sqrt(4-x)dx = 61/6 + 64/6 = 125/6 (Check!)

The second problem will work better with y(x) than x(y).