Help with double integration

1.

∫∫D x^2*y^2 dxdy

The D is the number of coordinates in the first quadrant that satisfies the inequality x+y ≤ 1

2.

∫∫D x^9*y^9

D = {(x,y) : x^4 ≤ y ≤ x^(1/5)

∫∫D x^2*y^3*ln(x^2+y^2) dxdy

D = {(x,y) : x^2 ≤ x^2 + y^2 ≤ x^4, (x,y) ≥ 0

I just wonder, how do I know between which intervals I should integrate.

Thanks

1) Write D as y = 0 to y = 1 - x with x in [0, 1].
(Sketching the region may be useful.)

So, we obtain
∫(x = 0 to 1) ∫(y = 0 to 1 - x) x^2 y^2 dy dx
= ∫(x = 0 to 1) x^2 * (1/3) y^3 {for y = 0 to 1 - x} dx
= (1/3) ∫(x = 0 to 1) x^2 (1 - x)^3 dx
= (1/3) ∫(x = 0 to 1) (x^2 - 3x^3 + 3x^4 - x^5) dx
= (1/3) (1/3 - 3/4 + 3/5 - 1/6)
= 1/180.
-----------
2) Since y = x^4 and y = x^(1/5) intersect at x = 0 and 1, we have
∫(x = 0 to 1) ∫(y = x^4 to x^(1/5)) x^9 y^9 dy dx.

This is straightforward to integrate.
-----------
3) Do you really mean D = {(x,y) : x^2 ≤ x^2 + y^2 ≤ x^4, (x,y) ≥ 0}?
This is a strange way to write a region of integration.

Okay; use polar coordinates for #3. 1 ≤ x^2 + y^2 ≤ 4 with (x,y) ≥ 0 ==> 0 ≤ r ≤ 1 and 0 ≤ θ ≤ π/2.
∫∫D x^2 y^3 ln(x^2+y^2) dx dy
= ∫(θ = 0 to π/2) ∫(r = 1 to 2) (r^2 cos^2(θ)) * 9r^3 sin^3(θ)) ln(r^2) * r dr dθ
= ∫(θ = 0 to π/2) ∫(r = 1 to 2) cos^2(θ) sin^3(θ)) * 18r^6 ln r dr dθ.

Try it now...

Report Abuse