- #1
Hixy
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I've been fooling around by myself with the book "div, grad, curl and all that" by H.M. Schey to learn some vector calculus. However, in the second chapter, when he performs the integrals, he skips the part where he finds the limits on x and y. Here's an example:
Compute the surface integral [tex]\int \int_{S} (x+y)dS[/tex]
where S is the portion of the plane x+y+z=1 in the first octant.
yada yada yada some rewriting etc.
The final integral is [tex]\sqrt{3}\int \int_{R} (1-y)dxdy[/tex], R being the area of S projected onto the xy-plane. He then says "this is a simple double integral with value 1/√3, as you should be able to verify.
What I take from this:
"In the first octant" means the first quadrant in the xy-plane, octant being used because of 3-dimensional space.
Finding the limits on x and y, I set z=y=0 to find x, and z=x=0 to find y, both limits being 0 to 1 from the x+y+z=1 equation. But I don't get the same value for the integral as the solution says. Where does it go wrong?
Compute the surface integral [tex]\int \int_{S} (x+y)dS[/tex]
where S is the portion of the plane x+y+z=1 in the first octant.
yada yada yada some rewriting etc.
The final integral is [tex]\sqrt{3}\int \int_{R} (1-y)dxdy[/tex], R being the area of S projected onto the xy-plane. He then says "this is a simple double integral with value 1/√3, as you should be able to verify.
What I take from this:
"In the first octant" means the first quadrant in the xy-plane, octant being used because of 3-dimensional space.
Finding the limits on x and y, I set z=y=0 to find x, and z=x=0 to find y, both limits being 0 to 1 from the x+y+z=1 equation. But I don't get the same value for the integral as the solution says. Where does it go wrong?