Evaluate the integral by reversing the order of integration: Z 1 0 Z 1 √ x p y 3 + 1 dy dx Sketch the domain of integration.

Question

Evaluate the integral by reversing the order of integration: Z 1 0 Z 1 √ x (y^3 + 1)^1/2 dy dx Sketch the domain of integration.

Answer 1

Change dydx to dxdy

If you draw the plot, the bounds of the inside integral change from √x to 1 to 0 to y²

because the trapped area is between y = 1 and y = √x or x=0 and x=y²

so the inside integral is now∫

∫(y³+1)^1/2dx = (y³+1)x evaluated from 0 to y² = y²(y³+1)

Now the outside integral is:

∫y²(y³+1)^1/2dy evaluated from 0 to 1

Let u = y³+1 , du = 3y²dy, when y = 0, u= 1, when y = 1, y= 2√

(1/3)∫√udu = (2/9)u^3/2 evaluated from 1 to 2 = (2/9)√8 - (2/9) = (2^5/2-2)/9 =0.40631