Question

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} y d y d x $$

Short answer

The region of integration is a semi-circle in the first quadrant. The value of the double integral is \(\frac{1}{3}\).

Step-by-step solution

Understand the region of integration

The region of integration is given by the bounds of the integral: \(y\) is bounded between 0 and \(\sqrt{1-x^2}\) and \(x\) is between 0 and 1. This represents a semi-circle in the first quadrant.

Evaluate the inner integral

The inner integral can be evaluated as: \[ \int_{0}^{\sqrt{1-x^{2}}} y d y = [\frac{y^2}{2}]_{0}^{\sqrt{1-x^{2}}} = \frac{1}{2} (1 - x^2) \] After substituting the limits of \(y\), we get a function in terms of \(x\).

Evaluate the outer integral

Now, integrate the result of inner integral with respect to \(x\): \[ \int_{0}^{1} \frac{1}{2}(1-x^{2}) dx = [\frac{x}{2} - \frac{x^3}{6}]_{0}^{1} = \frac{1}{3} \]