Question

If ∅ = 𝟐𝒙𝒚𝒛 𝟐 , 𝑭 = 𝒙𝒚𝒊 − 𝒚𝒋 + 𝒙 𝟐𝒌 𝐚𝐧𝐝 𝑪 𝐢𝐬 𝐭𝐡𝐞 𝐜𝐮𝐫𝐯𝐞 𝒙 = 𝒕 𝟐 , 𝒚 = 𝟐𝒕, 𝒛 = 𝒕 𝟑 𝐟𝐫𝐨𝐦 𝒕 = 𝟎 𝒕𝒐 𝒕 = 𝟏, 𝐞𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐭𝐡𝐞 𝐥𝐢𝐧𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬 a. ∫ ∅𝒅𝒓 𝑪 , b. ∫ 𝑭 𝑿 𝒅𝒓.

Answer 1

$\phi=2xyz^{2}$

$F=xy$$\hat{i}-y\hat{j}+x^{2}\hat{k}$

Curve C is given by $x=t^{2},y=2t,z=t^{3}$ from t=0 to t=1 .

We have to evaluate line integeral -

$a)\int_{c}\phi$ $dr$

$d\vec{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

$\phi$ =$2xyz^{2}$ , now making this variable in terms of t , we have to put the given $x,y\ and \ z$ so that we can get a function in t .

$x=t^{2}$ $,y=2t$ $,z=t^{3}$

$\phi=$ $2$ $t^{2}\times2t\times(t^{3})^{2}$

$\phi=4t^{9}$

$dx=2t$$dt$ $,$ $dy=2dt$ $,dz=3t^{2}dt$

$d\hat{r}=(2t\hat{i}+2\hat{j}+3t^{2}\hat{k})dt$

$=\int_c\phi$ $d\hat{r}=\int_{0}^{1}($ $2t\hat{i}+2\hat{j}+3t^{2}\hat{k})$ $.4t^{9}dt$

$=\int_0^1(8t^{10}\hat{i}+8t^{9}\hat{j}+12t^{11}\hat{k})dt$

$=\dfrac{8}{11}\hat{i}+\dfrac{8}{10}\hat{j}+t^{12}\hat{k}$

$B)$ $\int F.Xdr$

$d\hat{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

$d\hat{r}=(2t\hat{i}+2\hat{j}$ $+3t^{2}\hat{k})dt$

$F=xy\hat{i}-y\hat{j}+x^{2}\hat{k}$

now putting the value of $x, y ,z$ in above equation we get -

$F=$ $2t^{3}\hat{i}-2t\hat{j}+t^{4}\hat{k}$

$=\int_{0}^{1}(2t^{3}\hat{i}-2t\hat{j}+t^{4}\hat{k}).(2t\hat{i}+2\hat{j}+3t^{2}\hat{k})dt$

$=\int_{0}^{1}(4t^{4}-4t+3t^{6})dt$

$=\dfrac{4}{5}-2+\dfrac{3}{7}$

$=\dfrac{28-70+15}{35}=\dfrac{-27}{35}$