Find the line integrals of 𝐅 from (0,0,0) to (1,1,1) over each of the following paths in the acc

step 1 We have three parts. So first of all the path one we have if equals 3y i cap plus 2x j cap plus

Find the line integrals of 𝐅 from (0,0,0) to (1,1,1) over...

Find the line integrals of $\mathbf{F}$ from (0,0,0) to (1,1,1) over each of the following paths in the accompanying figure. a. The straight-line path $C_{1}: \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1$ b. The curved path $C_{2}: r(r)=r i+t^{2} j+r^{4} k, \quad 0 \leq t \leq 1$ c. The path $C_{3} \cup C_{4}$ consisting of the line segment from (0,0,0) to (1,1,0) followed by the segment from (1,1,0) to (1,1,1) (FIGURE CAN'T COPY) $$\mathbf{F}=3 y \mathbf{i}+2 x \mathbf{j}+4 z \mathbf{k}$$

Find the line integrals of $\mathbf{F}$ from (0,0,0) to (1,1,1) over each of the following paths in the accompanying figure.
a. The straight-line path $C_{1}: \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1$
b. The curved path $C_{2}: r(r)=r i+t^{2} j+r^{4} k, \quad 0 \leq t \leq 1$
c. The path $C_{3} \cup C_{4}$ consisting of the line segment from (0,0,0)
to (1,1,0) followed by the segment from (1,1,0) to (1,1,1)
(FIGURE CAN'T COPY)
$$\mathbf{F}=3 y \mathbf{i}+2 x \mathbf{j}+4 z \mathbf{k}$$

Key Concepts

Potential Function

A potential function is a scalar function whose gradient yields the original vector field in a conservative field. By finding a potential function, the evaluation of a line integral becomes straightforward, as it can be computed simply by subtracting the value of the potential function at the initial point from its value at the terminal point.

Path Independence

Path independence is a property of conservative vector fields whereby the value of a line integral between two points remains the same regardless of the curve taken. This fundamental concept in vector calculus allows one to choose the most convenient path for evaluation, often reducing the complexity of the integral.

Line Integrals

A line integral in vector calculus is a method of integrating a function along a curve. It is used to compute quantities such as work done by a force field along a path, requiring the evaluation of a vector field dotted with an infinitesimal displacement vector along the curve.

Parametrization

Parametrization is the process of expressing a curve or path using a parameter, typically denoted by t. This allows the coordinates of points on the curve to be written as functions of t, which facilitates the computation of line integrals by transforming them into integrals with respect to the parameter.

Conservative Vector Fields

A conservative vector field is one in which the line integral between any two points is independent of the path taken. This implies that the field can be expressed as the gradient of a scalar potential function, which simplifies the process of evaluating line integrals since they only depend on the values of the potential function at the endpoints.

Transcript

00:01 We have three parts.

00:01 So first of all the path one we have if equals 3y i cap plus 2x j cap plus 4 z k cap.

00:16 Now we have this c1 equal to r of t which is equal to t i cap plus t j cap plus t k cap.

00:33 Therefore we have d r over d t equals i plus j plus k now f of x y and z equal to f of t t and t so it is equal to 3 t i this is i plus 2 t j plus 4 t therefore integration of c1 f dot d r is equal to integration 0 to 1 f of r of t into d r over d t into d d so we have this as integration of 0 to 1 3 t i plus 2 t j plus 40 k into i plus i plus j plus k d t is this is given as addition of 0 to 1 and this is 3 t plus 2 t plus 40 into d this is equal to the addition of 0 to 1 90 d d this is equal to 9 upon 2 after putting the limits so we have this 9 by 2 which is the answer of part a of the problem we have part b of the problem where f x y z equal to f of t t square and t to the power of four is equal to three t squared i plus two t j plus four to the power four k now we have this c2 is equal to r of t is equal to t is equal to t i plus t squared j plus t to the power of 4k this is equal to now we have d r over d t this is equal to i plus 2 t j plus 40 to the power q k now we have to solve for c2 therefore integration of c2 f dot d r is given as integration 0 to 1 f r of t into d r over d t into d t so coming back to the integration part we have this equals to integration 0 to 1 3 t square i plus 2 t j plus 4 t is k into i plus 2dj plus 4d to the power 3 k into d now after solving this integration we have this as 13 upon 3 which is the answer for part b of the problem now we have part c of the problem where f of x y z equals to 3y 2x and 4 z so f of r of t is given as f of t t and 0 this is given as 3 t 2 t and 0 this is 3 t 2 t and 0 now we have c3 that is equal to r of t which is equal to t, t and zero...

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