Evaluate the l in integral ?_C ( x ^2 + y^ 2 ) d s along the path C: r ( t ) = ( c o s t + t...

Question:

Evaluate the lin integral {eq}\int_{C}(x^{2}+y^{2})ds {/eq} along the path C: {eq}r(t)=(cos \ t+t \ sin \ t)i+(sin \ t -t \ cos \ t)j,0\leq t \leq 2\pi {/eq}

Line integral of scalar fields:

Derivatives and integration both will be used here. 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}

Parametric equations are given by

{eq}r(t)=\langle \cos t+t\sin t, \sin t-t\cos t \rangle {/eq}

Derivative is

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

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

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

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

{eq}\displaystyle \int_C (x^2 + y^2) ds=\int_{0}^{2}[ (\cos t+t\sin t)^2+(\sin t-t\cos t)^2 ]\times t dt {/eq}

{eq}\displaystyle \int_C (x^2 + y^2) ds=\int_{0}^{2}(\cos^2 t+t^2 \sin^2 t+2t\sin t \cos t+\sin^2 t+t^2 \cos^2 t-2t\sin t \cos t )t dt {/eq}

{eq}\displaystyle \int_C (x^2 + y^2) ds=\int_{0}^{2} (1+t^2)t dt= \int_{0}^{2} (t+t^3) dt {/eq}

{eq}\displaystyle \int_C (x^2 + y^2) ds= [ \frac{t^2}{2}+\frac{t^4}{4} ]_{0}^{2} {/eq}

{eq}\displaystyle \int_C (x^2 + y^2) ds= 2+4 {/eq}

{eq}\displaystyle \int_C (x^2 + y^2) ds= 6 {/eq}


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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