Question

Evaluate the partial integral. $$ \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right) d x $$

Short answer

The value of the integral is \(\frac{2\left(1-y^{2}\right)\sqrt{1-y^{2}}}{3}\)

Step-by-step solution

Performing Integration

Initially, treat \(y\) as constant and perform the integration. Since we are integrating \(x^2 + y^2\) with respect to \(x\), we get \[\int (x^2 + y^2) dx = x^3/3 + y^2x.\]

Evaluate the integral between the limits

Now, evaluate this integral between the limits \(-\sqrt{1-y^{2}}\) and \(\sqrt{1-y^{2}}\).This yields \[\left[ x^3/3 + y^2x \right]_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} = \frac{\left(\sqrt{1-y^{2}}\right)^{3}}{3} + y^{2}\sqrt{1-y^{2}}-\left(\frac{\left(-\sqrt{1-y^{2}}\right)^{3}}{3}-y^{2}\sqrt{1-y^{2}}\right).\]

Simplification

Simplify this expression to get\[2\sqrt{1-y^{2}}\left(\frac{\left(1-y^{2}\right)}{3}+y^{2}\right)=\frac{2\left(1-y^{2}\right)\sqrt{1-y^{2}}}{3}.\]