I think that I found it,
you can reduce the big squareroot of death to a mere
\sqrt{1+\frac{-1}{-∞}} = 1 so the final equation is equal to
\int_{-1}^{1} 2\pi\sqrt\frac{x^2 (1 - x^2)}{8} * 1 dx
= 2\pi\int_{-1}^{1} \sqrt\frac{x^2 (1 - x^2)}{8}dx
which is a lot simplier to integrate
Hehe I know right the original function is
\sqrt{ \frac{x^2 (1 - x^2)}{8}}
so I have to do the surface of revolution formula on this
which is defined
S = \int_a^b2\pi f(x)\sqrt{1+(\frac{dy}{dx})^2}dx
Greeting everyone I am trying in integrate this function, to obtain the surface of revolution of a function.
1. Relevant equations
\int_{-1}^{1} 2\pi\sqrt\frac{x^2 (1 - x^2)}{8}\sqrt{1+(\left| x\right | \frac{-2x^2+1}{2(-x^2+1)^{1/2}*2^{1/2}x}})^2
2. The attempt at a solution
I tried to...
no I can't use the kinect equation, but i found the answer with calculus
f(0) = 32 so
speed = f(x) = -9.8x + 32
so if I integrate this from 0 to 32/9.8 it gives me 52.24 meters
(-9.8t^2)/2 + 32x
(-9.8(32/9.8)^2)/2 + 32(32/9.8) = 52.24 meters
which is a mere aproximate
Homework Statement
A ball is thrown at 32m/s vertically what is the max height the ball can attain.
Homework Equations
no equation is given except initial speed
The Attempt at a Solution
so my guess is since the gravity is 9.8m/s^2
if I integrate 9.8m/s^2 = 32 it gives me 2.1396...