Question

Let \({\rm{C}}\)be a simple closed smooth curve that lies in the plane \({\rm{x + y + z = 1}}\). Show that the line integral

\(\int_{\rm{C}} {{\rm{z dx - 2x dy + 3y dz}}} \)

depends only on the area of the region enclosed by\({\rm{C}}\) and not on the shape of \({\rm{C}}\)or its location in the plane.

Short answer

The line integral \(\int_{\rm{C}} {\rm{z}} {\rm{dx - 2xdy + 3ydz}}\)depends only on \(\int\limits_{\rm{C}} {{\rm{F}} \cdot {\rm{dr}}} {\rm{ = }}\frac{{{\rm{2A}}}}{{\sqrt {\rm{3}} }}\)(\({\rm{A}}\)is the area of the region enclosed by the smooth curve \({\rm{C}}\) ) and not on the shape of \({\rm{C}}\) or its location in the plane.

Step-by-step solution

Concept Introduction

An integral is a mathematical notion that explains displacement, area, volume, and other ideas that come from the combination of infinitesimal data. Integration is the term used to describe the process of locating integrals.

According to Stoke's theorem"The surface integral of the curl of a function over a surface limited by a closed surface is equivalent to the line integral of the particular vector function around that surface,".

Equation for Integral

Re-write the integral as –

\(\begin{array}{c}\int_{\rm{C}} {\rm{z}} {\rm{dx - 2xdy + 3ydz = }}\int_{\rm{C}} {{\rm{(zi - 2xj + 3yk)}}} \cdot {\rm{(dxi + dyj + dzk)}}\\{\rm{ = }}\int_{\rm{C}} {\rm{F}} \cdot {\rm{dr}}\end{array}\)

Now, the Stoke’s Theorem suggests –

\(\int_{\text{C}} {\text{F}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{curl F}}} \cdot {\text{dS}}\)

Where\({\rm{C}}\)is the boundary of the surface oriented counter-clockwise.

Choose the surface whose boundary is\({\rm{C}}\).

Let\({\rm{S}}\)be the part of the plane\({\rm{x + y + z = 1}}\), which is the region enclosed by the loop\({\rm{C}}\).

By the definition, it can be written –

\({\rm{curlF = }}\left( {\frac{{\partial {\rm{R}}}}{{\partial {\rm{y}}}}{\rm{ - }}\frac{{\partial {\rm{Q}}}}{{\partial {\rm{z}}}}} \right){\rm{i + }}\left( {\frac{{\partial {\rm{P}}}}{{\partial {\rm{z}}}}{\rm{ - }}\frac{{\partial {\rm{R}}}}{{\partial {\rm{x}}}}} \right){\rm{j + }}\left( {\frac{{\partial {\rm{Q}}}}{{\partial {\rm{x}}}}{\rm{ - }}\frac{{\partial {\rm{P}}}}{{\partial {\rm{y}}}}} \right){\rm{k}}\)

Note that\({\rm{F = z i - 2x j + 3y k}}\). Hence, it can be obtained –

\(\begin{array}{c}{\rm{curlF = (3 - 0)i + (1 - 0)j + ( - 2 - 0)k}}\\{\rm{curlF = 3i + j - 2k}}\end{array}\)

Finding Area of the Region

Now on further evaluating –

\(\int_{\text{C}} {\text{F}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{curl F}}} \cdot {\text{dS}}\)

Note that \({\rm{dS = ndS}}\).

\({\text{ = }}\iint_{\text{S}} {{\text{curl F}}} \cdot {\text{ndS}}\)

Since \({\rm{S}}\) is the part of the plane \({\rm{x + y + z = 1}}\). So, there is \({\rm{n = }}\frac{{\rm{1}}}{{\sqrt {\rm{3}} }}{\rm{(i + j + k)}}\).

Remember that \({\rm{ai + bj + ck}}\) is normal to the plane \({\rm{ax + by + cz = d}}\).

\(\begin{gathered} {\text{ = }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\iint_{\text{S}} {{\text{(3i + j - 2k)}}} \cdot {\text{(i + j + k)dS}} \\ {\text{ = }}\frac{{\text{1}}}{{\sqrt {\text{3}} }}\iint_{\text{S}} {\text{3}}{\text{ + 1 - 2dS}} \\ {\text{ = }}\frac{{\text{2}}}{{\sqrt {\text{3}} }}\iint_{\text{S}} {\text{d}}{\text{S}} \\ {\text{ = }}\frac{{{\text{2A}}}}{{\sqrt {\text{3}} }} \\ \end{gathered} \)

Note that \(\iint\limits_{\text{s}} {{\text{dS}}}\) is the area inside the loop \({\rm{C}}\), which is said to be \({\rm{A}}\).

Therefore, the line integraldepends only on the area of the region enclosed by\({\rm{C}}\).