anonymous
  • anonymous
having trouble getting started on hydrostatic pressure in calc II. I thought I had it down in class, but I am getting none of the answers right, so I must be forgetting some step. we have a shape /| | | |_| where the top triangle is 1m tall and wide, the bottom rectangle portion is 1m wide and 2m tall, the whole surface is suspended vertically in a liquid with mass density of 500kg/m^3. the topmost corner of the triangle portion is at the point of the surface. Calculate the force in newtons on the surface. I tried integrate(500*9.8*y^2dy) from 0 to 1 plus integrate(500*9.8*ydy) from
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1333427330281:dw|
amistre64
  • amistre64
i assume the bottom is easy enough to calculate
amistre64
  • amistre64
\[\int_{0}^{1}\int_{1}^{3}500\ dd.dw\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
the top part we just need to adjust our dd to include a function of w as a starting point
amistre64
  • amistre64
\[\int_{0}^{1}\int_{f(w)}^{1}500\ dd.dw\]
amistre64
  • amistre64
since it mimcs the points w=0, d=1; and w=1, d=0 we have a slope of -1 d = -w+1
amistre64
  • amistre64
of course it doesnt matter what way the top we position; we could just as well have turned it about such that w=0, d=0 and w=1 d=1 to get d = w
amistre64
  • amistre64
but then we would be measuring from 0 to f(w)
amistre64
  • amistre64
i spose you need the 9.8 to make things newtonian
anonymous
  • anonymous
\[\int\limits_{0}^{1}500*9.8*depth*width*dy\] would be \[\int\limits_{0}^{1}500*9.8*y^2dy\] for the top portion?
anonymous
  • anonymous
or is that too many y's? (depth y, plus width as a function of y right?
amistre64
  • amistre64
.... forget y and x; those are pointless; use variable you can relate to, d for depth and w for width i think are the simplest
amistre64
  • amistre64
|dw:1333428899517:dw| look for how you want to measure this thing; from what to what
amistre64
  • amistre64
d goes from 1 to 2 on the bottom; and w goes from 0 to 1 the top simplifies to d goes form w to 1; and w goes from 0 to 1
amistre64
  • amistre64
the top measures out as: \[(w)\int_{0}^{1}\left((d)\int_{w}^{1} dd\right)dw\] add that to the bottom \[(w)\int_{0}^{1}\left((d)\int_{1}^{2} dd\right)dw\]
amistre64
  • amistre64
since 500 and 9.6 are constants they pull out to till the end anyways
amistre64
  • amistre64
int dd = d from w to 1; 1-w int 1-w dw = w - w^2/2 from 0 to 1,: 0 is pointless so 1-1/2 = 1/2:*C
anonymous
  • anonymous
not used to seeing this as a double integral... have to get my mind around that.
amistre64
  • amistre64
the under is similar int dd = d from 1 to 2; 2 - 1 = 1 int 1 dw = w from 0 to 1 = 1. 1*C
amistre64
  • amistre64
so if we are lucky :) C(1/2+1) is our answer with C being that 500*9.6
amistre64
  • amistre64
7200 N ?
amistre64
  • amistre64
i write them as doubles so that i can see what im measuering to and not having to rely upon some sort of mental gymnastics to mess up on
amistre64
  • amistre64
which doesnt mean i dont mess up, its just more obvious as to why when i do :)
anonymous
  • anonymous
I must be just too old, double integrals make my head hurt... but let me run back through on paper to see if I understand it right. Calc II just gets harder the farther in I go....
amistre64
  • amistre64
you 40 yet?
anonymous
  • anonymous
returning student, 47 in 7 days.
amistre64
  • amistre64
:) well then, you got 7 years on me
anonymous
  • anonymous
yeah, but I thought I survived Calc I, but I at least took a D in it
anonymous
  • anonymous
27 years ago...
anonymous
  • anonymous
so I had an introduction.... this is all new....
amistre64
  • amistre64
lol, i aint got nothing less than an A on the maths yet
anonymous
  • anonymous
That D 27 years ago is why I dropped out of college... now I am back, and starting to wonder why...

Looking for something else?

Not the answer you are looking for? Search for more explanations.