Question

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{2} \int_{0}^{4-x^{2}} x y^{2} d y d x $$

Short answer

The result of the integral is \(-\frac{1}{3}\)

Step-by-step solution

- Represent the domain of integration graphically

The limits of \(x\) from 0 to 2 and \(y\) from 0 to \(4 - x^2\), give us a quarter circle region in the first quadrant of a Cartesian coordinate system, with radius 2 (as \(x^2 + y^2 = r^2\), ie., \(x^2 + (4 - x^2) = 2^2\))

- Solving the inner integral

Start by taking the inner integral with respect to \(y\), treating \(x\) as a constant throughout this process.\n\n\(\int_{0}^{4-x^{2}} x y^{2} d y = [\frac{1}{3}x*y^3]_{y=0}^{y=4-x^2} = \frac{1}{3}x*(4-x^{2})^3 - 0\)

- Solve the integral obtained in Step 2

To solve step 2, we substitute \(y = 4 - x^2\) into the equation.\n\n\[\frac{1}{3}x*(4-x^{2})^3 - 0\]

- Calculating the outer integral

Next, we evaluate the outer integral by treating the result at the end of step 3 as the integrand.\n\n\[\int_{0}^{2} (\frac{1}{3}x*(4-x^{2})^3) dx = [\frac{1}{12}*(4-x^2)^4]_{x=0}^{x=2} = \frac{1}{12}* (0 - 2^4)\] = \(-\frac{1}{3}\)

- Conclusion

The result of the double integral is \(-\frac{1}{3}\)