Question

Evaluate the definite integral \(\int_{\rm{0}}^{\rm{a}} {\rm{x}} \sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} {\rm{dx}}\)

Short answer

So, the evaluation of integration is \(\int_{\rm{0}}^{\rm{a}} {\rm{x}} \sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} {\rm{dx = }}\frac{{\rm{1}}}{{\rm{3}}}{{\rm{a}}^{\rm{3}}}\)

Step-by-step solution

Find\({\rm{u}}\).

So the,

\(\begin{array}{l}{\rm{u = }}{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}\\{\rm{du = - 2xdx}}\;\;\; \Rightarrow \;{\rm{ - }}\frac{{{\rm{du}}}}{{\rm{2}}}{\rm{ = xdx}}\end{array}\)

The new integration constraints of\({\rm{u}}\).

If \(\begin{array}{l}{\rm{u = 0 , if x = a}}\\{\rm{u = }}{a^2}{\rm{ ,if }}x = 0\end{array}\)

Substituting \({\rm{u}}\) to function\(_{\rm{0}}^{\rm{a}}{\rm{x}}\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} {\rm{dx}}\).

\(\begin{array}{c}{\rm{x}}\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} {\rm{dx = }}{{\rm{a}}^{\rm{2}}}\sqrt {\rm{u}} {\rm{ - }}\frac{{{\rm{du}}}}{{\rm{2}}}\\{\rm{ = - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}{{\rm{u}}^{{\rm{1/2}}}}{\rm{du}}\\{\rm{ = - }}{\left. {\frac{{\rm{1}}}{{\rm{2}}}\frac{{{{\rm{u}}^{\frac{{\rm{3}}}{{\rm{2}}}}}}}{{\frac{{\rm{3}}}{{\rm{2}}}}}} \right|_{{{\rm{a}}^{\rm{2}}}}}\\{\rm{ = - }}{\left. {\frac{{\rm{1}}}{{\rm{3}}}{{\rm{u}}^{\frac{{\rm{3}}}{{\rm{2}}}}}} \right|_{{{\rm{a}}^{\rm{2}}}}}\\{\rm{ = - }}\frac{{\rm{1}}}{{\rm{3}}}{{\rm{0}}^{\frac{{\rm{3}}}{{\rm{2}}}}}{\rm{ - }}{\left( {{{\rm{a}}^{\rm{2}}}} \right)^{\frac{{\rm{3}}}{{\rm{2}}}}}\\{\rm{ = - }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{ - }}{{\rm{a}}^{\rm{3}}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}{{\rm{a}}^{\rm{3}}}\end{array}\)

Hence the value\(\int_{\rm{0}}^{\rm{a}} {\rm{x}} \sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} {\rm{dx = }}\frac{{\rm{1}}}{{\rm{3}}}{{\rm{a}}^{\rm{3}}}\).