Calculus

Sorry to bother you so much but do you think you could check my answers for these two:

jiskha.com/questions/1788952/Hi-I-missed-school-due-to-illness-and-my-school-isnt-very-forgiving-so-I-would

jiskha.com/questions/1788953/I-know-it-looks-like-a-lot-but-its-just-10-questions-part-1-It-also-explains

thank you so much for all your help so far!

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asked by George
  1. I'd have to say they all look good to me...
    Mainly because I wrote those posts!

  2. Sorry - I didn't see your final answers.
    Everything looks good except as noted below
    1st set
    1D. Hmmm. Maybe I'm off. Better check your text for those definitions.
    2C. Things are easy for discs. Using shells, the boundary changes at y = 1/27
    Below there, the solid is just a cylinder of radius 1/27 and height 2
    Above there, the shells just belong to the small triangular curved region.
    I could have expressed the cylinder part as an integral, but it would just be
    ∫[0,1/27] 2πy*2 dy = 2πy^2 [0,1/27] which is the same thing as the cylinder formula.
    2nd set
    #4. Nope. 8 Use unsigned values, since we want actual distance traveled, not displacement.
    #5. Not a choice. Use unsigned area
    #7. Oops. Twice that. Can you spot my typo?
    #10. Come on. F = mg (weight)

  3. 1D. So it says that average acceleation is Δv/Δt so it would be (17-17)/(1.4-1) so 0/.4 which is 0. So, I think your right with the no net acceleration.
    2C. Do you only add this to the shell and not the disc? So I got that the volume of the cylinder is 0.0086189, how should I write this as a final answer?

    5. How do I find the unsigned area? (I'm sorry, this is probably a really dumb question)
    10. The only values given are t=5 and t=7. I'm sorry but I am a bit confused on how to finish solving p = ∫F dt

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    posted by George
  4. 2C The boundary for the discs is a single curve, so there is no extra volume to include. I only included the shells to show that the two methods produce the same answer.

    5. Since the graph is sometimes negative, simple integration subtracts the area below the x-axis. Figure each area geometrically, rather than algebraically.

    10. come on, guy. They told you that F(t) = cos t
    So, ∫[5,7] cost dt = 1.616

  5. Thank you so much for including both methods on 2C it was very intersting although I was a bit confused. Does that mean that the answer is just 242π/1215?

    Also, I was wondering why number 4 is 8? If we used unsigned values wouldn't it be 7? The area of the first triangle is 6 and the smaller on is 1, have I made a mistake?

    So for 5 am I supposed to ignore that the areas under the x-axis are negative? If I go off the triangle areas found in number 4 (area of 7) would It then be 7 + pi/2? (sorry If I'm botching a lot of this up)

    I'm so sorry I was so confused on number 10, for some reason I thought I had to have a value to substitute the t in cos t, and the first time, the way it's written on my thing made it look like cost and I was very confused.

    Also thank you so so much for all your help and your patience with me although I am probably really frustrating due to my lack of understanding, I promise you I appreciate this so much.

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    posted by George

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