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Now that we have setup the parameterization let's try to evaluate the integral.

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The next thing we need to do is express ds in terms of du and dv.

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We have seen this before ds is going to be equal to the magnitude of the cross product

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of partial of r with respect to u crossed with partial of r with respect to v du dv .

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First lets take the cross product and we will do that with a 3 by 3 matrix , we will do that right over here

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we set up and fill in r of u and r of v in the actual determinant over here. So first we have our components i , j , and k.

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First think about r sub u is. The partial of r with respect to u. Well it's i component

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is going to be 1 the partial of u with respect to u is 1 so it's i component is 1 , it's j component is going to be zero

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partial of v with respect to u is zero , v doesnt change with respect to u , so this is going to be 0 , partial of (there have

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to be brackets around here partial of this with respect to u is just going to be 1 again , this is the partial of v squared respect

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to u is just zero , so this is just 1 again and then r sub v the partial of r with respect to v , the i component is going to be zero

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j component is going to be 1 and the partial of u plus v squared is going to be 2v , so this is pretty straight forward determinant

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so lets try to evaluate , so i component i so we cross this column and this row , so this be 0 times 2v minus 1 times 1 , so essentially

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it is just going to be negative 1 times i so we are going to have negative i this is equal to negative i and then the j component

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we have to put a negative out front. And remember we have to check out board pattern.

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So cross out that row, that column.