metacogitans Posted September 8, 2015 Share Posted September 8, 2015 (edited) So as some here may know, I've been teaching myself calculus. I learned differentiation and finding tangent lines to curves using infinitesimals with a little help from my friend's calc book; then about a year later I finally grasped the fundamental theorem of calculus after someone pointed me to an illustration on wikipedia which showed that the addition of an infinitely thin rectangle in relation to a section of area under a curve ends up working out to differentiation, except giving you the function for the curve instead of a derivative, meaning that inverse differentiation is used for finding the area under a curve. Then after that, with much struggle and confusion trying to understand differential operators and things like "the chain rule" all of which I had skipped over, I was eventually able to figure out implicit differentiation and actually make sense of it - which was the last problem I asked at this forum: finding the area under circles. Now, I'm finally moving on to the big boy calculus and am attempting to take on multivariable calculus. I find it easiest to learn by asking a problem and watching how its solved, so here it is: The equation in question is z = x^2 + y^2 + 2x + 2y + 2 The graphed equation looks like a sort of circular parabola bowl. Now, what I want to find is the 2 dimensional area under that shape's curve between the x values 0 and -3 on the line y=2x Edit: fixed the problem; the problem I had up before was unsolvable This is the 3 dimensional function: And this is the line segment that I want to find the 2 dimensional area of underneath the curve of the 3d function: Any takers? I'm looking for how you found the solution; the solution itself isn't that important. Edited September 8, 2015 by metacogitans Link to comment Share on other sites More sharing options...
overtone Posted September 9, 2015 Share Posted September 9, 2015 I'm not sure I follow, but what's wrong with plugging 2x in for every appearance of y and integrating the single variable function from -3 to 0? What did I miss? Link to comment Share on other sites More sharing options...
metacogitans Posted September 9, 2015 Author Share Posted September 9, 2015 (edited) I'm not sure I follow, but what's wrong with plugging 2x in for every appearance of y and integrating the single variable function from -3 to 0? What did I miss? If you can express that section of the 3 dimensional shape in 2 dimensions, you can do this. But if you look at the third picture closely, there's not really an easy way to write a function for that curve without using multiple variables. I posted the problem in /r/mathematics and no one's solved it there yet either. That makes me assume there's a huge drop-off between Calc II and Calc III Edited September 9, 2015 by metacogitans Link to comment Share on other sites More sharing options...
metacogitans Posted September 19, 2015 Author Share Posted September 19, 2015 (edited) Someone on reddit gave a solution to the problem finally; unfortunately they wrote out how they found it entirely with differential/integtal operators, which aren't readable unless you already know how to do the problem being solved, so I still have no clue how to find line integrals. Here's the solution he posted: A = int_S |ds/du X ds/dv| dudv (parameterized area integral) The surface is defined by x in [-3,0], y=2x, z in [0,x2+y2+2x+2y+2]. Let u=x, v=z, and write s(x,z) = [x,2x,z] in terms of u and v to get s(u,v) = [u,2u,v]. ds/du = [1,2,0], ds/dv = [0,0,1], ds/du X ds/dv = [2,-1,0], |ds/du X ds/dv| = sqrt(5) u in [-3,0], v in [0, 5u2+6u+2] A = int{-3:0}{0:5u2+6u+2} sqrt(5) dvdu = sqrt(5)int{-3:0} 5u2+6u+2 du Edited September 19, 2015 by metacogitans Link to comment Share on other sites More sharing options...
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