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Let F(x, y, z)=xi+yj+zk. Evaluate the integral of F along each of the following paths.

\[c(t)=(t,t,t), \space 0 \le t \le 3 \space\]

The aim of this question is to find the Integration of the given function $F (x, y, z) =i+ yj +zk$ by first integrating $F (t, t, t) $ and then we will put the values of the limits given with the function.

The basic concept behind this question is the knowledge of integration, the limits of integration, derivates, and integration rules such as the product and quotient integration rules.

Expert Answer

Given function we have:

\[ F (x, y, z) = i + yj + zk\]

Here given integral $ F (x, y, z) = i + yj + zk $ is to be evaluated along each of the indicated paths:

\[ c ( t ) = ( t, t, t) \]

So the limit of the given paths $ c ( t ) $ is given by:

\[ c ( t ) = ( t, t, t ) | \space 0 \le t \le 3 \space \]

Now to solve the given function with integration, we have to identify the limits of integration carefully. As given the limits of integral $ c(t)$ vary from $0 $ to $3$ which can be represented as:

\[ = \int_{ 0 }^{ 3 } \]

To find out the value of the line integral $F $ we will take the derivative of:

\[ c( t ) = ( t, t, t ) | \space 0 \le t \le 3 \space\]

\[\dfrac{ dc }{ dt } = ( t, t, t )\]

As the derivative of the given path is taken with respect to $t $ so:

\[\dfrac{ dc }{ dt } = ( 1, 1, 1 )\]

\[=\int_{0}^{3} {F (t, t, t) } \times \dfrac{dc}{ dt} dt\]

Putting value of $ \dfrac{ dc }{ dt } $ in above equation, we get:

\[=\int_{0}^{3} {F (t, t, t) } \times ( 1, 1, 1 ) dt\]

\[=\int_{0}^{3} {3t } \times ({ 1, 1, 1 }) dt\]

\[=\int_{0}^{3} {3t }dt\]

\[=3 \left[ t \right]_{0}^{3}\]

\[= 3 \left[ \dfrac{ t^2 }{ 2 } \right]_{0}^{3} \]

Putting the limit of $t $ in the above equation:

\[= 3 \left[ \dfrac{ (3)^2 }{ 2 } – \dfrac{ (0)^2 }{ 2 } \right] \]

\[= 3 \left[ \dfrac{ (3)^2 }{ 2 } – \dfrac{ 0 }{ 2 } \right] \]

\[= 3 \left[ \dfrac{ (3)^2 }{ 2 } – 0 \right] \]

\[= 3 \left[ \dfrac{ 9 }{ 2 } \right] \]

\[= 3 \times \dfrac{ 9 }{ 2 } \]

\[= \dfrac{ 27 }{ 2 }\]

Numerical Result

Integral $F$ is evaluated along each path as:

\[= \dfrac{ 27 }{ 2 }\]

Example

Find out the value of the line integral $F(t, t, t)$ with paths:

\[c(t)={ t, t, t } , \space 0 \le t \le 2\]

Solution

\[=\int_{0}^{2}{F (t, t, t)} \times \dfrac{dc}{ dt}dt\]

\[=\int_{0}^{2} {F (t, t, t) } \times ({ 1, 1, 1 }) dt\]

\[=\int_{0}^{2} {3t } \times ({ 1, 1, 1 })dt\]

\[=\int_{0}^{2} {3t }dt\]

\[=3\left[t\right]_{0}^{2}\]

\[=3\left[\dfrac{t^2}{2}\right]_{0}^{2}\]

\[=3\left[\dfrac{2^2}{ 2} – \dfrac{0^2}{ 2}\right]\]

\[=3\left[\dfrac{4}{ 2}\right]\]

\[=6\]

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