计算二重积分∬_D4cos²(2x+y)dxdy的变量替换求解咨询

Great question! Choosing (u=2x+y) is a perfect first step here—since the integrand is a function of (2x+y) alone, this substitution will simplify things a lot. Let's figure out the right (v) to pair with it.

First, recall that for a valid variable substitution in double integrals, we need a pair ((u, v)) that lets us rewrite (x) and (y) as functions of (u) and (v), with a non-zero Jacobian determinant. Since you've already fixed (u = 2x + y), we can pick a simple linear variable for (v) that makes inverting the system and mapping the original region (D) straightforward.

Option 1: Let (v = x) (my preferred choice for simplicity) #

If we set (v = x), we can solve for (x) and (y) in terms of (u) and (v) directly:

• (x = v)
• (y = u - 2v)

Now let's map the original region (D) to the (uv)-plane:
Original (D) is defined by:

• (2 ≤ 2x + y ≤ 6) → translates to (2 ≤ u ≤ 6)
• (x ≥ 0) → translates to (v ≥ 0)
• (y ≥ 0) → (u - 2v ≥ 0) → translates to (v ≤ u/2)

The (uv)-region is a trapezoidal area bounded by (u=2), (u=6), (v=0), and (v=u/2)—super easy to integrate over!

Calculate the Jacobian Determinant #

The Jacobian matrix for this substitution is:

∂(x,y)   | ∂x/∂u  ∂x/∂v |   | 0   1 |
-------- = |          | = |       |
∂(u,v)   | ∂y/∂u  ∂y/∂v |   | 1  -2 |

The determinant is ((0)(-2) - (1)(1) = -1). We take the absolute value for the integral: (|J| = 1).

Rewrite the Integral #

The integrand becomes (4\cos^2(u)) (since (2x+y = u)), so the integral transforms to:
$$
\int_{u=2}^{6} \int_{v=0}^{u/2} 4\cos^2(u) \cdot |J| , dv , du
$$

First integrate with respect to (v):
$$
\int_{0}^{u/2} dv = \frac{u}{2}
$$
So the integral simplifies to:
$$
\int_{2}^{6} 4\cos^2(u) \cdot \frac{u}{2} , du = \int_{2}^{6} 2u\cos^2(u) , du
$$

Evaluate the Single Integral #

Use the trigonometric identity (\cos^2(u) = \frac{1 + \cos(2u)}{2}) to rewrite the integrand:
$$
2u \cdot \frac{1 + \cos(2u)}{2} = u + u\cos(2u)
$$

Split the integral into two parts:

1. (\int_{2}^{6} u , du = \frac{1}{2}u^2 \bigg|_{2}^{6} = \frac{1}{2}(36 - 4) = 16)
2. (\int_{2}^{6} u\cos(2u) , du) (use integration by parts: let (t = u), (dt = du); (dv = \cos(2u)du), (v = \frac{1}{2}\sin(2u))):
   $$
   \frac{1}{2}u\sin(2u) \bigg|{2}^{6} - \int{2}^{6} \frac{1}{2}\sin(2u) , du
   $$
   $$
   = \left(3\sin(12) - \sin(4)\right) + \frac{1}{4}\cos(2u) \bigg|_{2}^{6}
   $$
   $$
   = 3\sin(12) - \sin(4) + \frac{1}{4}(\cos(12) - \cos(4))
   $$

Add the two results together for the final integral value:
$$
16 + 3\sin(12) - \sin(4) + \frac{1}{4}(\cos(12) - \cos(4))
$$

Option 2: Let (v = y) (another valid choice) #

If you prefer, you could also set (v = y). Then:

• (x = \frac{u - v}{2})
• (y = v)

The Jacobian determinant here is:

∂(x,y)   | 1/2  -1/2 |
-------- = |          | = (1/2)(1) - (-1/2)(0) = 1/2
∂(u,v)   | 0     1   |

The (uv)-region would be:

• (2 ≤ u ≤ 6)
• (v ≥ 0) (since (y ≥ 0))
• (x ≥ 0) → (\frac{u - v}{2} ≥ 0) → (v ≤ u)

The integral becomes:
$$
\int_{2}^{6} \int_{0}^{u} 4\cos^2(u) \cdot \frac{1}{2} , dv , du = \int_{2}^{6} 2u\cos^2(u) , du
$$
Which is exactly the same single integral as before—so you'll end up with the same result.

Key Takeaway #

When choosing (v), pick a variable that makes inverting the substitution (solving for (x) and (y)) easy, and maps the original region to a simple, bounded area in the (uv)-plane. (v = x) or (v = y) both work here, but (v = x) gives a Jacobian with absolute value 1, which saves a tiny bit of calculation.

内容的提问来源于stack exchange，提问作者Parseval