Use an integral test:
\int 1 / (n*ln|n|)
A certain substitution should eventually lead to an antiderivative of:
ln(ln(n))
I'll leave the rest to you (remember to look at what lim n->inf: ln(ln(n)) does)
Thanks, but I'm having trouble with the second integral in the problem:
2pg \int (y-1) * \sqrt{9-y^2}
For: 2pg\int \sqrt{9-y^2} from 1 - 3 (when integrals are seperated)
doesn't that evaluate to 2pg \int9*cos^2\theta ?
I'm not getting the correct answer when I evaluate this integral.
Homework Statement
Find the hydrostatic force on any side of a bottom half-circle with a 6 m diameter with the top 1 m above water level.
http://img377.imageshack.us/my.php?image=45095885io3.png"
Homework Equations
P = 1000 * 9.8 (mass density of water) * displacement
A = l * w
F = P * AThe...
Yeah, first year momentum-impulse problems usually don't account for gravity.
For a:
I = \Delta p, p = m * v
For b:
I = F*\Delta t, I = \Delta p,
F = \Delta p / \Delta t
ABG and DEH are congruent
GBC and HEF are congruent
angle ABG = angle DEH
angle GBC = angle HEF
AGB + GBC = DEH + HEF
ABC is congruent with DEF
Can you give reasons why this is?
(Hint: the first is given, the second has to do with properties of...
I don't know that it will be traveling at 10 m/s at the instance of the catch . . .
Actually, if it follows a perfectly horizontal path, I guess it is safe to assume.
Mentallic is correct, but I would just like to add that the first boy should be removed from your thought process. You have a ball traveling at you (50 kg person) with a certain momentum (mass * velocity). It dosen't matter how it got its momentum at this point. The momentum when the ball is...
I'd just like to add that you can think of the bullet and the leopard as an inelastic collision.
By the way, this is kind of a gruesome physics question :biggrin:
Sorry, I'm just getting used to all the quirks of latex.
The question is to find the length using surface area revoultion forumla of y = sin (pi * x) (rotated about the x-axis) from 0 - 1.
\int2\pi*sin(\pi*x) * \sqrt{1 + (sin(\pi*x)')^2}
I think sec^3 is actually correct. You have to...
Homework Statement
y = sin \pix Using arc length and surface revoultion on x-axis 0 <= x <= 1
The Attempt at a Solution
d/dx sin \pix = \pi cos \pix
(\pi cos\pix)^2 = \pi^2 cos^2\pix
\int sin pi * x * 2 * pi * \sqrt{1 + pi^2 * cos^2 (pi*x)}
u = pi cos (pi * x)
du = -pi^2 * sin...