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    Early introduction to calculus, for beginning-algebra students

    hint; max ht is when velocity=0. v=dh/dt. and as said previous you are missing a t. h(t) = 5 + 20t - 5t^2
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    derive eqn for vol of sphere using polar coordinates.

    derive eqn for vol of sphere using polar coordinates. Vol of sphere =4/3(pi)r3 to derive this eqn using polar coordinates Vs=int r2drd@sin&d& limits are r=0,r; @=0,2pi; &=0,pi were does the d& come from? i.e. there is no dimension d& i.e. in deriving eqn for Area of circle, Acircle=int rdrd@...
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    integrate Area of a rt triangle w/ hypotenuse=r, horizontal=x, vertical=sqrt(r2-x2)

    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&)...
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    integral [from x to 0] (r^2 - x^2)^{1/2} dx : different units

    int(r2-x2)1/2dx, limits x=x, 0 (x is upper limit, 0 is lower limit) =int r(1-x2/r2)1/2dx sub x/r=sin& =r2int cos2&d&, limits &=sin-1(x/r), 0 ?? cos2&=1/2(1+cos2&) =r2/2 int(1+cos2&)d& sub u=2& 1/2du=d& =r2/4 int(1+cosu)du limits u=2sin-1(x/r),0 ?? =r2/4 (u+sinu) evaluated at limits...
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    area of semi-circle, negative answer

    I am getting a negative number for the area of a semicircle, and I can't find my mistake. find: area of semi-circle of radius 2, centered at origin. A=S-22S0sq rt(4-x2)dydx, [i.e. limits of x are -2 and 2, limits of y are 0 and square root (4-x2)] 1st int. A=S-22sq rt (4-x2)dx change...