Question

evaluate $int_{c} f ds$ for the function $f(x,y)=sqrt{1 + 9xy}$ over the curve $y = x^{3}$ for $0leq xleq4$.
the value of the integral is

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Explanation:

Step1: Find the derivative of \(y = x^{3}\)

The derivative \(y^{\prime}=\frac{dy}{dx}=3x^{2}\).

Step2: Use the formula for the line - integral \(\int_{C}f(x,y)ds=\int_{a}^{b}f(x,y)\sqrt{1 + (\frac{dy}{dx})^{2}}dx\)

Here, \(f(x,y)=\sqrt{1 + 9xy}\), \(y = x^{3}\), so \(f(x,x^{3})=\sqrt{1+9x\cdot x^{3}}=\sqrt{1 + 9x^{4}}\), and \(\sqrt{1+(\frac{dy}{dx})^{2}}=\sqrt{1+(3x^{2})^{2}}=\sqrt{1 + 9x^{4}}\). Then \(\int_{C}fds=\int_{0}^{4}\sqrt{1 + 9x^{4}}\cdot\sqrt{1 + 9x^{4}}dx=\int_{0}^{4}(1 + 9x^{4})dx\).

Step3: Integrate term - by - term

\(\int_{0}^{4}(1 + 9x^{4})dx=\int_{0}^{4}1dx+\int_{0}^{4}9x^{4}dx\).
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\int_{0}^{4}1dx=x\big|_{0}^{4}=4-0 = 4\) and \(\int_{0}^{4}9x^{4}dx=9\cdot\frac{x^{5}}{5}\big|_{0}^{4}=\frac{9}{5}(4^{5}-0)=\frac{9}{5}\times1024=\frac{9216}{5}\).

Step4: Sum the results

\(\int_{0}^{4}(1 + 9x^{4})dx=4+\frac{9216}{5}=\frac{20 + 9216}{5}=\frac{9236}{5}=1847.2\).

Answer:

\(1847.2\)