Find the value of the double integral
x*(sqrt(x^2 + y^2)) dy dx
Note that the limit of x goes from 0 to 1 and the limit of y goes from o to x
To evaluate this double integral, we first need to integrate with respect to y and then with respect to x, using the given limits:
∫∫x*(sqrt(x^2 + y^2)) dy dx = ∫[0,1]∫[0,x]x*(sqrt(x^2 + y^2)) dy dx
Integrating with respect to y first:
∫[0,1]∫[0,x]x*(sqrt(x^2 + y^2)) dy dx = ∫[0,1]x∫[0,x](sqrt(x^2 + y^2)) dy dx
Using the u-substitution u = x^2 + y^2, so that du/dy = 2y:
∫[0,1]x∫[0,x](sqrt(x^2 + y^2)) dy dx = ∫[0,1]x∫[x^2,x^2 + x^2](1/2)*(sqrt(u)) (1/2)*du/dy dx
Simplifying the limits and pulling out constants:
= (1/2)∫[0,1]x∫[x^2,2x^2]sqrt(u) du dx
Integrating with respect to u:
= (1/2)∫[0,1]x[(2/3)*u^(3/2)]|[x^2,2x^2] dx
= (1/2)∫[0,1]x[(2/3)*(2x^3 - x^3)] dx
= (1/2)∫[0,1]x*(4/3)*x^3 dx
= (1/2)*(4/12)*x^5|[0,1]
= (1/2)*(4/12)*1
= 1/6
Therefore, the value of the double integral is 1/6.