- #1
Saketh
- 261
- 2
I'm working from H.M. Schey's Div, grad, curl, and all that, and am trying to figure out surface integration.
One of the example problems boils down to the following surface integral over a projection, with z = 1 - x - y
[tex]
\sqrt{3} \int \!\!\! \int_R 1 - y \,dx \,dy
[/tex]
I made x and y go from 0 to 1.
[tex]
\sqrt{3} \int_0^1 \!\!\! \int_0^1 1 - y \,dx \,dy
[/tex]
According to the book, the answer is [tex]\frac{1}{\sqrt{3}}[/tex], but I keep getting [tex]\frac{\sqrt{3}}{2}[/tex].
Another example problem has z = 1 - y - x/2:
[tex]
\int \!\!\! \int_R \frac{3x}{4} - \frac{3y}{2} + \frac{1}{2} \,dx \,dy
[/tex]
I made x go from 0 to 2, and y go from 0 to 1.
[tex]
\int_0^1 \!\!\! \int_0^2 \frac{3x}{4} - \frac{3y}{2} + \frac{1}{2} \,dx \,dy
[/tex]
The answer is supposed to be 1/2, but I keep getting 1.
I have not completely lost faith, because I am able to evaluate the integrals which require conversion to polar coordinates. However, the ones in cartesian coordinates are confusing me.
Thanks in advance.
One of the example problems boils down to the following surface integral over a projection, with z = 1 - x - y
[tex]
\sqrt{3} \int \!\!\! \int_R 1 - y \,dx \,dy
[/tex]
I made x and y go from 0 to 1.
[tex]
\sqrt{3} \int_0^1 \!\!\! \int_0^1 1 - y \,dx \,dy
[/tex]
According to the book, the answer is [tex]\frac{1}{\sqrt{3}}[/tex], but I keep getting [tex]\frac{\sqrt{3}}{2}[/tex].
Another example problem has z = 1 - y - x/2:
[tex]
\int \!\!\! \int_R \frac{3x}{4} - \frac{3y}{2} + \frac{1}{2} \,dx \,dy
[/tex]
I made x go from 0 to 2, and y go from 0 to 1.
[tex]
\int_0^1 \!\!\! \int_0^2 \frac{3x}{4} - \frac{3y}{2} + \frac{1}{2} \,dx \,dy
[/tex]
The answer is supposed to be 1/2, but I keep getting 1.
I have not completely lost faith, because I am able to evaluate the integrals which require conversion to polar coordinates. However, the ones in cartesian coordinates are confusing me.
Thanks in advance.