Math Problem Statement

∫01​∫x^2,2-x^2 (xy))​dydx reverse order and solve

Solution

To reverse the order of integration and solve the double integral:

I=01x22x2xydydxI = \int_0^1 \int_{x^2}^{2-x^2} xy \, dy \, dx

Step 1: Identify the region of integration

We first need to understand the limits of the integral. The inner integral is with respect to yy, and it ranges from y=x2y = x^2 to y=2x2y = 2 - x^2. The outer integral is with respect to xx, and it ranges from x=0x = 0 to x=1x = 1.

So the region of integration RR is defined by: x2y2x2and0x1x^2 \leq y \leq 2 - x^2 \quad \text{and} \quad 0 \leq x \leq 1

Step 2: Reverse the order of integration

Now we want to switch the order of integration, i.e., integrate with respect to xx first, then with respect to yy.

From the bounds, we can express the region as:

  • For a fixed yy, the values of xx range from x=0x = 0 to x=yx = \sqrt{y} (since y=x2y = x^2) and from x=2yx = \sqrt{2 - y} to x=1x = 1 (since y=2x2y = 2 - x^2).

Thus, the new integral becomes: I=010yxydxdy+012y1xydxdyI = \int_0^1 \int_0^{\sqrt{y}} xy \, dx \, dy + \int_0^1 \int_{\sqrt{2-y}}^1 xy \, dx \, dy

Step 3: Solve the integral

First part: 010yxydxdy\int_0^1 \int_0^{\sqrt{y}} xy \, dx \, dy

  • Integrate with respect to xx: 0yxydx=y2x20y=y2y=y22\int_0^{\sqrt{y}} xy \, dx = \frac{y}{2} x^2 \Bigg|_0^{\sqrt{y}} = \frac{y}{2} \cdot y = \frac{y^2}{2}
  • Now, integrate with respect to yy: 01y22dy=12y3301=16\int_0^1 \frac{y^2}{2} \, dy = \frac{1}{2} \cdot \frac{y^3}{3} \Bigg|_0^1 = \frac{1}{6}

Second part: 012y1xydxdy\int_0^1 \int_{\sqrt{2-y}}^1 xy \, dx \, dy

  • Integrate with respect to xx: 2y1xydx=y2[x2]2y1=y2(1(2y))=y2y=y22\int_{\sqrt{2-y}}^1 xy \, dx = \frac{y}{2} \left[ x^2 \right]_{\sqrt{2-y}}^1 = \frac{y}{2} \left( 1 - (2 - y) \right) = \frac{y}{2} \cdot y = \frac{y^2}{2}
  • Now, integrate with respect to yy: 01y22dy=16\int_0^1 \frac{y^2}{2} \, dy = \frac{1}{6}

Step 4: Final result

Adding both parts together: I=16+16=13I = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}

Thus, the value of the double integral is 13\frac{1}{3}.

Would you like a detailed explanation on any step?

Here are 5 related questions:

  1. How do you determine the region of integration from given bounds?
  2. What techniques can be used to reverse the order of integration for non-rectangular regions?
  3. How can substitution help simplify double integrals?
  4. What are common applications of double integrals in physics or engineering?
  5. How does changing the order of integration affect the limits?

Tip: Always sketch the region of integration when reversing the order of integration for clarity.

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Math Problem Analysis

Mathematical Concepts

Double Integrals
Reversing Order of Integration
Integration in Rectangular Regions

Formulas

∫∫_R f(x,y) dy dx
Integral bounds transformation: From ∫_0^1∫_x^2^(2-x^2) xy dydx to ∫_0^1 ∫_0^√y xy dxdy

Theorems

Fubini's Theorem
Basic Integration Theorem

Suitable Grade Level

Undergraduate Calculus