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Natural sciences course of the University of Cambridge

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Year 1 of the natural sciences course of the University of Cambridge

Mathematics course of the year 1 of the natural sciences course of the University of Cambridge

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge

2004

Paper 2

Take the <dot product> of the given field with the upward <vector area element>. The surface relation simplifies its integrand:
$$
\mathbf F\cdot d\mathbf S=[y(x+z)+x]dx\,dy=(1-x^3+x)dx\,dy.
$$
The projected domain is exactly the rectangle $1/2\leq x,y\leq1$, since $y>0$ determines a unique $z=(1-x^3)/y-x$. Therefore the <surface integral> is
$$
\boxed{\int_{S_1}\mathbf F\cdot d\mathbf S=\frac12\int_{1/2}^1(1-x^3+x)dx=\frac12\left[x-\frac{x^4}{4}+\frac{x^2}{2}\right]_{1/2}^1=\frac{41}{128}.}
$$

Solution

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 7 / a / iii / Solution

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 7 / a / iii / Solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 7 / a / iii / Solution

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