I'm working from H.M. Schey's(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Double Integral

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