Area of a rt triangle, with the hypotenuse=r, horizontal=x, and vertical=sqrt(r2-x2)
I should get an equivalent to A=1/2base*ht
A=int sqrt (r2-x2)dx, limits for x=0,x
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&, limits 0, rsin&
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&, limits 0, rsin(1/2u)
A=(r2/4) int (1+cosu) du;
integrate and evaluate @ limits 0, rsin(1/2u)
=(r2/4) {[rsin1/2u+sin(rsin1/2u)]-[0+0]}
=(r2/4) {[rsin&+sin(rsin&)]-[0+0]}
questions;
1) I end up with r3, for an area??
2) what do i do with sin(rsin&)??
I should get an equivalent to A=1/2base*ht
A=int sqrt (r2-x2)dx, limits for x=0,x
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&, limits 0, rsin&
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&, limits 0, rsin(1/2u)
A=(r2/4) int (1+cosu) du;
integrate and evaluate @ limits 0, rsin(1/2u)
=(r2/4) {[rsin1/2u+sin(rsin1/2u)]-[0+0]}
=(r2/4) {[rsin&+sin(rsin&)]-[0+0]}
questions;
1) I end up with r3, for an area??
2) what do i do with sin(rsin&)??