Got Another Calculus Problem for You Guys; This Time Its Multivariable Integrals

[metacogitans]

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

[overtone]

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?

[metacogitans]

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

[metacogitans]

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