Question

In Exercises \(1-10\), evaluate the integral. $$ \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right) d x $$

Short answer

The definite integral of \(x^{2}+y^{2}\) from \(-\sqrt{1-y^{2}}\) to \(\sqrt{1-y^{2}}\) is \(\frac{2*(1 - y^{2})^{\frac{3}{2}}}{3}\).

Step-by-step solution

Calculate the Antiderivative of the Function

The antiderivative of \((x^{2}+y^{2})\) is \(\frac{x^3}{3} + xy^2\).

Substitute the Limits of Integration

Substitute the limits of integration into the antiderivative. Substituting \(\sqrt{1-y^{2}}\) for x gives \(\frac{(\sqrt{1-y^{2}})^3}{3} + \sqrt{1-y^{2}}*y^2\), and substituting \(-\sqrt{1-y^{2}}\) for x gives \(\frac{(-\sqrt{1-y^{2}}^3)}{3} - \sqrt{1-y^{2}}*y^2\).

Subtract to Obtain the Result

Subtract the result obtained when \(-\sqrt{1 - y^2}\) is substituted into the antiderivative from the result obtained when \(\sqrt{1 - y^2}\) is substituted into the antiderivative. This gives the definite integral as \( \frac{2*(1 - y^{2})^{\frac{3}{2}}}{3} \).