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Question 27.5: Evaluate the integral ∫C F(x, y) • ds, where F(x, y) = (3x²y......

Evaluate the integral \int_C \mathbf{F}(x, y) \cdot \mathrm{d} \mathbf{s} \text {, where } \mathbf{F}(x, y)=\left(3 x^2+y\right) \mathbf{i}+(5 x-y) \mathbf{j} and C is the portion of the curve y = 2x² between A(2, 8) and B(3, 18).

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\int_C \mathbf{F}(x, y) \cdot \mathrm{d} \mathbf{s}=\int_C\left(3 x^2+y\right) \mathrm{d} x+(5 x-y) \mathrm{d} y

The curve C has equation y=2 x^2 \text { for } 2 \leqslant x \leqslant 3. Along this curve we can replace y by 2 x^2 \text {. Note also that } \frac{\mathrm{d} y}{\mathrm{~d} x}=4 x and so we can replace dv with 4x dx. This will produce an integral entirely in terms of the variable x whieh is then integrated between x = 2 and x = 3. Thus

\begin{aligned} \int_C\left(3 x^2+y\right) \mathrm{d} x+(5 x-y) \mathrm{d} y & =\int_2^3\left(3 x^2+2 x^2\right) \mathrm{d} x+\left(5 x-2 x^2\right)(4 x \mathrm{~d} x)\end{aligned}

\begin{aligned}=\int_2^3 25 x^2-8 x^3 \mathrm{~d} x\end{aligned}

\begin{aligned} & =\left[\frac{25 x^3}{3}-2 x^4\right]_2^3 \\ & =28.333 \end{aligned}

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