Math Problem Statement

0101y2x2+y2\differentialDxdy\int_0^1\int_0^{\sqrt{1-y^2}}\sqrt{x^2+y^2}\differentialD\text{xdy}

Solution

The given double integral is:

0101y2x2+y2dxdy\int_0^1 \int_0^{\sqrt{1-y^2}} \sqrt{x^2 + y^2} \, dx \, dy

This represents the integral of x2+y2\sqrt{x^2 + y^2} over a region bounded by the curve x=0x = 0 to x=1y2x = \sqrt{1 - y^2}, and yy ranging from 0 to 1. Let's evaluate this step by step.

Step 1: Understand the Region of Integration

The region described by the limits is part of a quarter circle of radius 1 in the first quadrant.

Step 2: Change to Polar Coordinates

To simplify the calculation, let's convert to polar coordinates:

  • x=rcosθx = r \cos \theta
  • y=rsinθy = r \sin \theta
  • The Jacobian of the transformation is rr.

The limits in polar coordinates are:

  • rr ranges from 0 to 1.
  • θ\theta ranges from 0 to π2\frac{\pi}{2}.

The integrand x2+y2\sqrt{x^2 + y^2} in polar coordinates simplifies to rr.

Step 3: Rewrite the Integral in Polar Coordinates

0π201rrdrdθ=0π201r2drdθ\int_0^{\frac{\pi}{2}} \int_0^1 r \cdot r \, dr \, d\theta = \int_0^{\frac{\pi}{2}} \int_0^1 r^2 \, dr \, d\theta

Step 4: Evaluate the Inner Integral with respect to rr

01r2dr=[r33]01=13\int_0^1 r^2 \, dr = \left[\frac{r^3}{3}\right]_0^1 = \frac{1}{3}

Step 5: Evaluate the Outer Integral with respect to θ\theta

0π213dθ=13[θ]0π2=13π2=π6\int_0^{\frac{\pi}{2}} \frac{1}{3} \, d\theta = \frac{1}{3} \left[\theta\right]_0^{\frac{\pi}{2}} = \frac{1}{3} \cdot \frac{\pi}{2} = \frac{\pi}{6}

Thus, the value of the integral is:

π6\boxed{\frac{\pi}{6}}

Would you like more details on any of these steps or have any other questions?

Related Questions:

  1. How do you determine the region of integration for double integrals?
  2. What are the steps to convert Cartesian coordinates to polar coordinates in integration?
  3. How can you evaluate integrals involving x2+y2\sqrt{x^2 + y^2} without polar coordinates?
  4. What are the benefits of using polar coordinates in double integration?
  5. How does the Jacobian affect the transformation of integrals?

Tip: For regions with circular symmetry, converting to polar coordinates can often simplify the integration process significantly!

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Math Problem Analysis

Mathematical Concepts

Double Integrals
Cartesian Coordinates
Polar Coordinates

Formulas

Polar coordinates transformation formula
Jacobian determinant in polar coordinates

Theorems

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Suitable Grade Level

Advanced Undergraduate