# Evaluate \int_C G \cdot d r , where G = \langle 2xy, x^2 \rangle and C : r(t) = \langle...

## Question:

Evaluate {eq}\int_C \mathbf G \cdot d\mathbf r {/eq}, where {eq}\mathbf G = \langle 2xy, x^2 \rangle {/eq} and {eq}C : \mathbf r(t) = \langle t, 1-t\rangle ;\quad 0 \leq t \leq 1 {/eq}

## Line integral across a curve :

Derivatives have important applications. The line integral of a vector function across a parametric curve C is evaluated with the help of a dot product and

a derivative. First put the values from the parametric function {eq}r(t) {/eq} into {eq}F(x,y,z) {/eq} and then take the dot product with {eq}r'(t) {/eq}. Finally integrate it with respect to t with given range of t as the limits of integration.

{eq}\mathbf G = \langle 2xy, x^2 \rangle {/eq}

Parametric curve is given by

{eq}r(t)= \langle t,1-t \rangle {/eq}

So {eq}x=t, y=1-t {/eq}

Now writing the function G(x,y) in parametric form

{eq}G(r(t))=\langle 2t(1-t), t^2 \rangle {/eq}

{eq}G(r(t))=\langle 2t-2t^2, t^2 \rangle {/eq}

Find the derivative of parametric curve

{eq}r'(t)=\langle 1,-1 \rangle {/eq}

In vector calculus, line integral are evaluated as

{eq}\int_C \mathbf G \cdot d\mathbf r=\int_{a}^{b} F(r(t))\cdot r'(t) dt {/eq}

As {eq}0\leq t\leq 1, a=0, b=1 {/eq}

{eq}\int_C \mathbf G \cdot d\mathbf r=\int_{0}^{1} \langle 2t-2t^2,t^2 \rangle \cdot \langle 1,-1 \rangle dt {/eq}

{eq}\int_C \mathbf G \cdot d\mathbf r=\int_{0}^{1} [ 2t-2t^2-t^2 ] dt {/eq}

{eq}\int_C \mathbf G \cdot d\mathbf r=\int_{0}^{1} (2t-3t^2)dt {/eq}

{eq}\int_C \mathbf G \cdot d\mathbf r=[ t^2-t^3 ]_{0}^{1} {/eq}

{eq}\int_C \mathbf G \cdot d\mathbf r=1-1 {/eq}

{eq}\int_C \mathbf G \cdot d\mathbf r=0 {/eq}