Simple Nature - Light and Matter

∫ avolume ==∫ az=0 y=0∫ a ∫ a= a= a= a 3z=0 y=0∫ a ∫ az=0∫ az=0y=0a dz∫ ax=0a dy dzdy dzdx dy dzArea of a circle example 20⊲ Find the area of a circle.⊲ To make it easy, let’s find the area of a semicircle and thendouble it. Let the circle’s radius be r, and let it be centered on theorigin and bounded below by the x axis. Then the curved edgeis given by the equation r 2 = x 2 + y 2 , or y = √ r 2 − x 2 . Sincethe y integral’s limit depends on x, the x integral has to be on theoutside. The area isarea ==Substituting u = x/r,= r∫ rx=−r∫ rx=−r∫ rx=−rarea = r 2 ∫ 1∫ √ r 2 −x 2dy dxu=−1y=0√r 2 − x 2 dx√1 − (x/r) 2 dx .√1 − u 2 duThe definite integral equals π, as you can find using a trig substitutionor simply by looking it up in a table, and the result is, asexpected, πr 2 /2 for the area of the semicircle.4.2.5 Finding moments of inertia by integrationWhen calculating the moment of inertia of an ordinary-sized objectwith perhaps 10 26 atoms, it would be impossible to do an actualsum over atoms, even with the world’s fastest supercomputer. Calculus,however, offers a tool, the integral, for breaking a sum downto infinitely many small parts. If we don’t worry about the existenceof atoms, then we can use an integral to compute a momentSection 4.2 Rigid-Body Rotation 273