Integrate the function f at the path that consists of the curves C1 and C2

In summary, the conversation discusses the integration of a function along a specific path consisting of two curves. The solution involves breaking it into two separate integrals and then finding the sum of the two. The final answer is simplified to $\dfrac{5\sqrt{5}+9}{6}$, and the other person confirms that it is correct.
  • #1
mathmari
Gold Member
MHB
5,049
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Hey! :eek:

I have the following exercise:
Integrate the function $f=x+ \sqrt{y} -z^2$ at the path from $(0,0,0)$ to$(1,1,1)$ that consists of the curves $C_1:r(t)=t \hat{ \imath}+t^2\hat{ \jmath}, 0 \leq t \leq 1$ and $C_2: r(t)=\hat{ \imath}+\hat{ \jmath}+t \hat{k}, 0\leq t \leq1$.My solution is as followed:
$$\int_C{f}ds=\int_{C_1}{f}ds+\int_{C_2}{f}ds$$
$$\int_{C_1}{f}ds=\int_0^1{(t+\sqrt{t^2})\sqrt{1+4t^2}}dt=\int_0^1{2t(1+4t^2)^{\frac{1}{2}}}dt=\frac{1}{6}(1+4t^2)^{\frac{3}{2}}|_0^1=\frac{1}{6}(5 \sqrt{5}-1)$$
$$\int_{C_2}{f}ds=\int_0^1(2-t^2) \cdot 1 dt=[2t-\frac{t^3}{3}]_0^1=\frac{5}{3}$$
So $$\int_C{f}ds=\frac{5 \sqrt{5}-1}{6}+\frac{5}{3}$$Could you tell me if this is correct??
 
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  • #2
mathmari said:
Hey! :eek:

I have the following exercise:
Integrate the function $f=x+ \sqrt{y} -z^2$ at the path from $(0,0,0)$ to$(1,1,1)$ that consists of the curves $C_1:r(t)=t \hat{ \imath}+t^2\hat{ \jmath}, 0 \leq t \leq 1$ and $C_2: r(t)=\hat{ \imath}+\hat{ \jmath}+t \hat{k}, 0\leq t \leq1$.My solution is as followed:
$$\int_C{f}ds=\int_{C_1}{f}ds+\int_{C_2}{f}ds$$
$$\int_{C_1}{f}ds=\int_0^1{(t+\sqrt{t^2})\sqrt{1+4t^2}}dt=\int_0^1{2t(1+4t^2)^{\frac{1}{2}}}dt=\frac{1}{6}(1+4t^2)^{\frac{3}{2}}|_0^1=\frac{1}{6}(5 \sqrt{5}-1)$$
$$\int_{C_2}{f}ds=\int_0^1(2-t^2) \cdot 1 dt=[2t-\frac{t^3}{3}]_0^1=\frac{5}{3}$$
So $$\int_C{f}ds=\frac{5 \sqrt{5}-1}{6}+\frac{5}{3}$$Could you tell me if this is correct??

It looks good to me! (Yes)

I'd probably further simplify it to $\dfrac{5\sqrt{5}+9}{6}$, but other than that, I see nothing wrong with your work. (Smile)
 
  • #3
Chris L T521 said:
It looks good to me! (Yes)

I'd probably further simplify it to $\dfrac{5\sqrt{5}+9}{6}$, but other than that, I see nothing wrong with your work. (Smile)

Great! Thank you very much! (Mmm)
 

FAQ: Integrate the function f at the path that consists of the curves C1 and C2

What is the purpose of integrating a function along a path?

The purpose of integrating a function along a path is to calculate the total area under the curve of the function over that specific path. This can be used to solve various problems in physics, engineering, and other fields.

What is the difference between integrating a function along a straight line and along a curved path?

When integrating a function along a straight line, the path is well-defined and can be easily calculated using basic geometry. However, when integrating along a curved path, the calculation becomes more complex as the path is not a simple straight line.

How do you determine the direction of integration along a path?

The direction of integration along a path is typically determined by the orientation of the path itself. For example, if the path is a closed loop, the direction of integration is usually counterclockwise. However, if the path is a line segment, the direction of integration can be from either end of the segment.

Can the function f be integrated along any type of path?

Yes, in theory, any function can be integrated along any type of path. However, in practice, the complexity of the path may make the integration process too difficult or impossible to solve analytically. In these cases, numerical methods may be used to approximate the solution.

What is the relationship between the curves C1 and C2 when integrating a function f along their path?

The relationship between the curves C1 and C2 depends on the specific function being integrated. In some cases, the curves may intersect or overlap, while in others, they may be completely separate. The function f may also have different values or behave differently along each curve, which can affect the overall integration result.

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