Question

why would I have to use an integral to compute work done instead of using the formula W=Force * distance?

Answer 1

Suppose the force varies and the path you do the work over isn't in a straight line? You might have trouble.

Answer 2

Would line integrals work? (I haven't used this before actually)

Answer 3

WHAT!? You can't be Feynman if you've never done a line integral.

Answer 4

I'm not Feynman. I'm his fan. :P

Answer 5

what exactly is meant by a constant force?

Answer 6

@cstarxq A force is constant if it doesn't change.

Answer 7

A force that doesn't vary over time. Like gravity, for the most part, can be considered a constant force. But suppose you have a kid kicking a wall over and over again? He exerts a force, then stops. Then starts again with the next kick.

Answer 8

@Kainui would that be considered a constant force? If a force is exerted, then stopped, and re-exerted...

Answer 9

okay, just making sure. Thank you so much everyone!

Answer 10

no calculus II (for no other reason than to kill myself)

Answer 11

thanks! I'll try to spend more time doing the hw problems and understanding the materials, but this site is amazingly helpful

Answer 12

Yeah, there's some powerful and fascinating stuff to come out of Calculus II and it's worth it to really understand what an integral is. Too many people stop at being comfortable being able to compute them... But to understand and apply them is powerful and unlocks an amazing super power of calculus.

For instance, have you ever noticed that the derivative of:
\[A=\pi r^2\]
is \[C=2\pi r\]

That comes from understanding integrals. (and derivatives.)

Answer 13

I think they should mention that in Cal 1 honestly. I mention this all the time to people who've taken up to differential equations and still haven't heard of this amazing use.

Answer 14

wow, yeah. I kinda skipped a year in between taking calc I and II so there are some basic stuff I need to polish up on.@redbullx thanks, see ya!

Answer 15

See, you start with the definition of pi.
\[\pi = \frac{ C }{ d }\] and forget 3.14. It's the ratio of a circle's circumference to its diameter. It doesn't matter what value it has, all that matters to us is that it is a constant and doesn't change depending on how large the circle is.

From here you can then imagine taking a circle and filling it with an infinite number of circles of smaller radius. Afterall, between two points you can pick an infinite number of smaller points right? Now if you were to give all those circles an infinitesimal small width they'd each have area. Add them all up from radius = 0 to radius of the circle you have and you have what you always wanted... pi*r^2.

Answer 16

Now try to derive the volume of a sphere using this new formula we created and you can use the pythagorean theorem to help you out. Shell/washer ftw.

Answer 17

Then if you get brave, derive volume of a donut. Also once you get volume of a sphere, take the derivative with respect to r and you get the surface area of the sphere. B)

Answer 18

kewl

Answer 19

Yeah, lotsa fun, you should definitely think about it, maybe consider deriving other similar things, like how about a sphere that can deform like this:



Like a waterballoon with a constant volume. Then find the surface area. =P