# Evaluate the line intergral along the curve C \int_{C} xy^{3} ds, where C is given by x = 4 \sin...

## Question:

Evaluate the line intergral along the curve C

{eq}\displaystyle \hspace{1cm} \int_{C} xy^{3} ds, \hspace{0.25cm} {/eq} where {eq}\displaystyle C {/eq} is given by {eq}\displaystyle x = 4 \sin t, y = 4 \cos t, z = 3t, 0 \leq t \leq \frac{\pi}{2} {/eq}

## Line Integral of Scalar Fields:

Derivatives have many applications. A derivative of a position vector will be used to solve this integral. First of all, evaluate the value of the function {eq}F(x,y,z) {/eq} in terms of the given parametric curve {eq}r(t) {/eq}. Then, multiply it with the magnitude of its derivative, i.e. {eq}\left | r'(t) \right | {/eq}, and then integrate it with the given limits of t. So, {eq}\displaystyle \int_C F\ ds=\int_{a}^{b} F(r(t)) \left | r'(t) \right | dt {/eq}

## Answer and Explanation:

{eq}\displaystyle \int_C F\ ds=\int_{a}^{b} F(r(t)) \left | r'(t) \right | dt {/eq}

{eq}r(t)=\langle 4\sin t, 4\cos t,3t \rangle {/eq}

Take the derivative

{eq}r'(t)=\langle 4\cos t,-4 \sin t, 3 \rangle {/eq}

{eq}\left | r'(t) \right |=\sqrt{16\cos^2 t+16\sin^2 t+9}=\sqrt{16+9}=\sqrt{25}=5 {/eq}

{eq}0\leq t\leq \frac{\pi}{2} {/eq}

{eq}\displaystyle \int_C (xy^3) \ ds=\int_{0}^{\pi/2}(256 \sin t \cos^3 t)\times 5 dt {/eq}

{eq}\displaystyle \int_C (xy^3) \ ds=1280 \int_{0}^{\pi/2} \sin t \cos^3 t dt {/eq}

{eq}\displaystyle \int_C (xy^3) \ ds=-1280 [ \frac{\cos^4 t}{4} ]_{0}^{\pi/2} {/eq}

{eq}\displaystyle \int_C (xy^3) \ ds= -320 [ \cos^4 t ]_{0}^{\pi/2} {/eq}

{eq}\displaystyle \int_C (xy^3) \ ds= -320(-1)=320 {/eq}