Area Between Curves

Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to c or y, Then find the area of the region,

2y+ 3sqrtx, y = 3, and 2y + 3 = 6

Answer 1

Alternatively to William's solution, we could integrate with respect to y instead of with respect to x. Doing it that way, we would need only one integral instead of two:

If 2y = 3√x, then x = (4/9)y².

If 2y + 3x = 6, then x = 2 - (2/3)y

Within the region, y varies from 1.5 to 3 (see William's diagram)

Area = ∫_{(1.5 to 3)} [(4/9)y² - (2 -(2/3)y)]dy = [(4/27)y³ - 2y + (1/3)y²]_{(1.5 to 3)}

= [(4 - 6 + 3] - [ 1/2 - 3 + 3/4] = 1 - (-1.75) = 2.75

Answer 2

If the equations are supposed to be:

1) 2y = 3√x (which is the same as y = 1.5√x)

2) y = 3

3) 2y + 3x = 6 (which is the same as y = -1.5x + 3)

Then here is the graph:

To find the area, you could break the integral into 2 and integrate as follows:

₀∫¹ (equation 2) - (equation 3) dx + ₁∫⁴ (equation 2) - (equation 1) dx

₀∫¹ (3) - (-1.5x + 3) dx + ₁∫⁴ (3) - (1.5√x) dx

₀∫¹ 3 + 1.5x - 3 dx + ₁∫⁴ 3 - 1.5√x dx

₀∫¹ 1.5x dx + ₁∫⁴ 3 - 1.5(x)^1/2 dx

(1.5/2)x²(evaluated between 0 & 1) + [3x - x^3/2]evaluated between 1 & 4

=[(3/4)(1²) - (3/4)(0²) + [3(4) - (4)^3/2] - [3(1) - (1)^3/2]

=3/4 + (12 - 8) - (3 - 1)

=3/4 + 4 - 2

=2.75