Question

The value of the line integral

\(\mathop \smallint \limits_c^\; \left( {2x{y^2}dx + 2{x^2} y dy + dz} \right)\)

along a path joining the origin  and the point (1,1,1)  is

Answer 1

Correct Answer - Option 2 : 2

Concept:

When two points (x₁, y₁. z₁) and (x₁, y₁. z₂) are mentioned find the relation in terms of the third variable in terms of x,y, and z:

\(\dfrac{{{{x}} - x_1}}{{x_2 - x_1}} = \dfrac{{{{y}} - y_1}}{{y_2 - y_1}} = \dfrac{{{{z}} - z_1}}{{z_2 - z_1}} = {{t}}\)

Put the value of z,y, and z and use the end-points of one variable.

Calculation:

Given:

\({\rm{I}} = \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right)\), A (0, 0, 0) and B(1, 1, 1).

Equation of line i.e. path

\( \frac{{{\rm{x}} - 0}}{{1 - 0}} = \frac{{{\rm{y}} - 0}}{{1 - 0}} = \frac{{{\rm{z}} - 0}}{{1 - 0}} = {\rm{t}} \)

\( \therefore {\rm{\;x\;}} = {\rm{\;y\;}} = {\rm{\;z\;}} = {\rm{\;t\;and\;t\;}}:{\rm{\;}}0{\rm{\;}} \to {\rm{\;}}1\)

\( \therefore {\rm{I}} = \mathop \smallint \limits_0^1 \left( {2{{\rm{t}}^3}{\rm{dt}} + 2{{\rm{t}}^3}{\rm{dt}} + {\rm{dt}}} \right) \)

\(= 4 \cdot \left[ {\frac{{{{\rm{t}}^4}}}{4}} \right]_0^1 + \left[ {\rm{t}} \right]_0^1 = 1 + 1 = 2\)

\( \therefore \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right) = 2 \)