Evaluate the line integral of f(x,y) along the curve C.
-$$f ( x , y ) = x , C : y = x ^ { 2 } , 0 \leq x \leq \frac { \sqrt { 35 } } { 2 }$$
A) $\frac { 215 } { 12 }$
B) $\frac { 215 } { 3 }$
C) $\frac { 215 } { 8 }$
D) 215

Question

Quizplus · Accepted Answer

The line integral of $f(x, y) = x$ along the curve $C: y = x^2$ from $0$ to $\frac{\sqrt{35}}{2}$ can be computed as follows:$$ \int_C f(x, y) \, ds = \int_{0}^{\frac{\sqrt{35}}{2}} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$Given $y = x^2$, we have $\frac{dy}{dx} = 2x$. Thus, the integral becomes:$$ \int_{0}^{\frac{\sqrt{35}}{2}} x \sqrt{1 + (2x)^2} \, dx = \int_{0}^{\frac{\sqrt{35}}{2}} x \sqrt{1 + 4x^2} \, dx $$Let $u = 1 + 4x^2$, so $du = 8x dx$, and when $x = 0$, $u = 1$, and when $x = \frac{\sqrt{35}}{2}$, $u = 1 + 4(\frac{\sqrt{35}}{2})^2 = 36$. Thus, the integral becomes:$$ \frac{1}{8} \int_{1}^{36} \sqrt{u} \, du = \frac{1}{8} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{36} = \frac{1}{12} \left[ 36^{3/2} - 1 \right] $$$$ = \frac{1}{12} \left[ 216 - 1 \right] = \frac{215}{12} $$Therefore, the correct answer is $\frac{215}{12}$, which is choice A.

Evaluate the Line Integral of F(x,y) Along the Curve C f(x,y)=x,C:y=x2,0≤x≤352f ( x , y ) = x , C : y = x ^ { 2 } , 0 \leq x \leq \frac { \sqrt { 35 } } { 2 }f(x,y)=x,C:y=x2,0≤x≤235​​

Evaluate the Line Integral of F(x,y) Along the Curve C $f ( x , y ) = x , C : y = x ^ { 2 } , 0 \leq x \leq \frac { \sqrt { 35 } } { 2 }$