# Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1029: 25

$$\sqrt 6$$

#### Work Step by Step

We have $r_u=1+j \space and \\ r_v=1-j+k$ and $$|r_u \times r_v|=\sqrt 6$$ $$Surface \space Area=\iint_S |r_u \times r_v| \space du \space dv \\=\int_0^1 \int_0^1 \sqrt 6 \space du \space dv \\=\sqrt 6(1-0) \\=\sqrt 6$$

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