Recent content by m0gh

After talking with the old teacher/friend I am here: -2x + 4 = (A+B+D)x^{3} + (-A +C - 2D + E)x^{2} + (A + B + D -2E)x -A + C + E Still working towards the answer. Please let me know if you see any mistakes

Just talked to an old teacher/friend and she said it should be set up as A/x-1 + (Bx+C)/(x-1)^(2) + (Cx + D)/(x^(2) + 1) so I'm not sure what you mean, Steam King

Homework Statement \int \frac{-2x + 4}{(x-1)^{(2)}(x^{(2)}+1)}Homework Equations The Attempt at a Solution I've done the problem a couple times but the answers keep coming out differently so I'm assuming I am messing up the setup. This is what I have for the first part of the setup: -2x +...

Okay, now I understand this problem and what I learned here has spilled over into other integration problems. I want to ask though. When you integrate along the x-axis are the rings still in the same position as integrating along the y-axis? When I think about integrating along x my mind...

I can definitely see the bigger picture now. The $$dl=\sqrt{dr^2 + (y(r+dr)-y(r))^2}\\$$ is the length of the line between (x, f(x)) and (x+dx, f(x+dx) from the Pythagorean theorem. Then using that as the "height" or "width" of the truncated cone and multiplying it by 2##\pi##r then...

I really appreciate your help with this, Simon. When you say: When you say the annulus between y and y+dy I am imagining the small change in the y is accounting for the very small height of the rectangle or rings. This height multiplied by length (2\pir) accounts for the surface area of...

Thanks everyone for your help with this one. I talked with my professor last night and her first question was why I set it up as x = \sqrt{9-y} She told me to try it again the way it is already set up. Honestly I did it the first way because it was my intuition and I didn't question it...

Can anyone spot a/the mistake?

I'm really sorry. I just forgot to multiply the pi and that's why I thought it was wrong. Thank you, Tanya. You have been very helpful and patient. I really enjoy when I ask for help and someone actually helps me to understand instead of just spitting out the answer.

Even using the (30-y) yields the same result after integrating. I think I'm more confused now than I was before.

Thank you so much. I took the integral from [0,30] of ∫51.2*100*pi*y*dy which gave: 5120∏[(y^2)/2] from [0,30] 5120∏[450-0] = 2,304,000∏ I hope this is correct I feel a lot better about these problems if so. EDIT: This answer is not correct. I just searched for this problem online...

Oh I think I see. If it is displaced y units then it should be weight*y ?

I'm sorry the wording of these problem types always gets me confused. So if the weight W = 100∏p*r^2*dy Then the work is determined by the integral of the weight from 0 to y?

If the volume of one disc is 100∏dy then I'm guessing (30-y)*100p∏dy would equal the weight?

If you mean a cross section that creates a circular disc then the Volume would be V = ∏(r2)dy Substitute 10 for r which gives V=∏ 100 dy?