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I am revisiting calculus after a number of years (24+!) and have stumbled on one question about hydro-static pressure. The original problem I solved was a triangle ABC, with vertical plane, below water. The top edge AB is 4 feet wide and is submerged 1 foot below the water, while C is 5 feet below. I basically set up the integral as 4/5w Integral(0 to 5) (h+1) (5-h) dh. With w as 62.5 lb/ft3, this came to 1,666 +2/3. Now, the followup question simply said flip the triangle 180 degrees, so that AB is still 1 foot below the water and C 4 feet above. I have struggled over this, with what I think of as the 'depth scaling' term, or the (h+1) term above, as the issue. I can't fathom what term to use to represent the under water bit of the triangle. So, any help will be greatly appreciated!

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    See if example 1 on this site http://tutorial.math.lamar.edu/Classes/CalcII/HydrostaticPressure.aspx can help you. It is very similar to your problem. – 2017-02-20
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    See also http://math.stackexchange.com/questions/159642/hydrostatic-pressure-on-an-equilateral-triangle?rq=1 and http://math.stackexchange.com/questions/157567/hydrostatic-pressure-on-a-square?noredirect=1&lq=1 – 2017-02-20
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    I divided the trapezoid base into a parallelogram and a triangle and used 'similar triangle' calcs to find width L, to wit, (H/(16/5)) = (L-(16/5)/(4/5)). This gives a width of L = H/4+16/5. The integral looks like, with W as weight of water: w Int(0 to 1) h (H/4+16/5)dh. This give me ~ 105+, where I'm supposed to get 111+2/3? – 2017-02-21
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    Btw, thanks for coming back. – 2017-02-21
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    Sorry, my mistake. The answer s/b 116+2/3. – 2017-02-21
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    Sorted it. Re-visited example at '...lamar...' and figured it out. Thanks for the steer. – 2017-02-22
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    Sorry, I was busy. Have you understood ? – 2017-02-22
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    Hi. Well, sort of... I followed, I believe, exactly the steps in 'Lamar', prob #2, to solve the prob. However, I did another very similar prob using this same method, a triangle with base = 4, height = 3, with the water 1 foot below the apex. The 'b' quantity in Lamar I find as b/(2-x) = (2/3)/2, and b = 1/3(2-x). I find the width of a representative rectangle as a = 4-2b. I calc this as a = 4-4/3+2/3x, or 8/3+2/3x. I set up the integral as w/3 Int(0 to 2) (8x+2/3x)dx. This gives me a wrong answer, which s/b/ 111+1/9. – 2017-02-23
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    2nd part: I note that by leaving out the 1st term, ie the 8x, I have the correct answer, ie 16/9w. Frankly, I can't figure it out as I have followed the steps as before! As always, any help is appreciated as I am a home study and, unfortunately, this is the 1 area where I don't have any other books, etc, to consult. Cheers. – 2017-02-23

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