Recent content by riemannsigma

I am having a difficult time seeing the three dimensional surface formed from a plane of equation 2x + y + z = 2 strictly inside the first quadrant. On the 2 dimensional xy plane, the closed, simple, piece wise curve is C1 along the x-axis from x=0 to 1, C2 along the line y= 2-2x is between...

Thanks

Let's say there is a 5 sided cube that is missing the bottom face. Obviously, there is a boundary curve at the middle of this cube that goes around the 4 sides, front, right, back, and left. This boundary curve forms the boundary of the top half of the cube with the 5 faces and the bottom...

You should try finding the surface integral using two parameters theta and z. Try taking the cross product to find the vector that points in direction of gradient. It's that easy

Try deriving the inertia of an inverted isosceles triangle using single variable calculus.

Surface area when using single variable calculus without parametrization requires the use of Lagrange's mean value theorem. So, instead of using dy, there is a need to use the infinitesimal ds.

Yes! Lol. Thanks

Also... The same inconsistency occurs for this conservative vector field. <e^y , xe^y> Having e in the parametric equation seems to make the line integral path DEPENDENT. WHATS GOING ON??

X axis- y-axis - z axis pathway. 1. x-axis y,z,dy,dz=0 Interval is from x=0 to x=x1 Integrate <0, 1, 0> • <dx, 0, 0> to get 0 2. Y-axis x=x1, dx,dz,z=0 Interval is from y=0 to y=y1 Integrate <y^2, 2x1y + 1, 3y> • <0, dy, 0> to get x1y^2 + y 3. Z-axis x=x1, y=y1, dx=dy=0 Interval is...

< y^2, 2xy+ e^(3z), 3ye^(3z)> is the vector field. the above vector field is inside an open simply connected domain. the parametric equations all have a continuous first order derivative inside the domain. Lastly, the curl of the vector field is <0, 0, 0> Thus, the vector field is...

Not continuous