Evaluate the double integral ∬Dx^2dA,
where D={(x,y):1≤x≤e,0≤y≤lnx
To evaluate the double integral ∬Dx^2dA over the region D={(x,y):1≤x≤e,0≤y≤lnx}, we first need to set up the limits of integration.
The region D is defined as the set of points (x, y) where 1≤x≤e and 0≤y≤lnx. In terms of a rectangular coordinate system, this means that D is a rectangle in the first quadrant, bounded by the x-axis, the line y=lnx, and the vertical lines x=1 and x=e.
We can express the limits of integration as follows:
∬Dx^2dA = ∫[a,b]∫[c,d]x^2dydx
where the limits of integration [a,b] and [c,d] are determined by the boundaries of the region D.
In this case, the boundaries of D are:
- The x-axis (y=0) as the lower boundary for y
- The line y=lnx as the upper boundary for y
- The vertical line x=1 as the lower boundary for x
- The vertical line x=e as the upper boundary for x
Therefore, the limits of integration are:
a = 1
b = e
c = 0
d = lnx
Putting it all together, we can express the double integral as:
∬Dx^2dA = ∫[1,e]∫[0,lnx]x^2dydx
Now, we can proceed to evaluate the double integral.
To evaluate the double integral ∬Dx^2dA over the region D={(x,y):1≤x≤e,0≤y≤lnx}, we can follow these steps:
Step 1: Set up the integral:
∬Dx^2dA = ∫₀ˡⁿˣ ∫₁ᵉ x^2 dxdy
Step 2: Evaluate the inner integral with respect to x:
∫₁ᵉ x^2 dx = [x^3/3] from 1 to e
= (e^3/3) - (1/3)
Step 3: Substitute the result back into the original integral:
∫₀ˡⁿˣ ∫₁ᵉ x^2 dxdy = ∫₀ˡⁿˣ (e^3/3 - 1/3) dy
Step 4: Evaluate the outer integral with respect to y:
∫₀ˡⁿˣ (e^3/3 - 1/3) dy = (e^3/3 - 1/3) [y] from 0 to lnx
= (e^3/3 - 1/3) (lnx - 0)
= (e^3/3 - 1/3) lnx
Therefore, the value of the double integral ∬Dx^2dA over the given region D is (e^3/3 - 1/3) lnx.