Help with Integral: Find the area of C

In summary, the conversation is about a specific integral involving an elliptic spiral and how to write it in a mathematical notation. The participants discuss the correct form of the integral and how to solve it. The correct form is determined to be \int_C [b*sin(t)*dx-a*cos(t)*dy+dz], and the participants discuss the values of dx, dy, and dz in this equation. They also discuss separating the integral and ensuring that the correct values are used in the equation.
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  • #2
so have you tried anything?
 
  • #3
here's how you can write it in tex
[tex] \int_C y dx-x dy +dz
[/tex]

note you can re-write as a dot product, just for clarity
[tex] \int_C (y,-x,1) \bullet \vec{dx} [/tex]

to solve, you just need to re-write the integral so you're only integrating over t
 
  • #4
lanedance said:
here's how you can write it in tex
[tex] \int_C y dx-x dy +dz
[/tex]

note you can re-write as a dot product, just for clarity
[tex] \int_C (y,-x,1) \bullet \vec{dx} [/tex]

to solve, you just need to re-write the integral so you're only integrating over t

So the integral becomes [tex] \int_C [/tex] [b*sint*dx-a*cost*dy+dz]
Which equals [tex] \int_C [/tex] -a*b*cost*sint*dt-a*b*cost*cost*dt+c*dt ? And now should I separate the integrals?

Is that all I have to solve or is something wrong?
 
  • #5
lanedance said:
here's how you can write it in tex
[tex] \int_C y dx-x dy +dz
[/tex]

note you can re-write as a dot product, just for clarity
[tex] \int_C (y,-x,1) \bullet \vec{dx} [/tex]

to solve, you just need to re-write the integral so you're only integrating over t

Or, a little neater write the integrand

[tex]\langle y, -x, 1\rangle \cdot \langle dx,dy,dz\rangle [/tex]
 
  • #6
gipc said:
So the integral becomes [tex] \int_C [/tex] [b*sint*dx-a*cost*dy+dz]
Which equals [tex] \int_C [/tex] -a*b*cost*sint*dt-a*b*cost*cost*dt+c*dt ? And now should I separate the integrals?

Is that all I have to solve or is something wrong?

so... is it okay to follow this steps?
 
  • #7
gipc said:
So the integral becomes [tex] \int_C [/tex] [b*sint*dx-a*cost*dy+dz]
Yes, that is correct.

Which equals [tex] \int_C [/tex] -a*b*cost*sint*dt-a*b*cost*cost*dt+c*dt ?
No, that is not correct. You are given that x= a cos t so dx= -a sin(t). b sin t dx= (b sin t)(-a sin t dt)= -ab sin^2 t dt. y= b sin t so dy= b cos t dt. -a cos t dy= (-a cos t)(b cos t dt)= -ab cos^2 t dt. z= ct so dz= ct as you have.

And now should I separate the integrals?

Is that all I have to solve or is something wrong?
 

1. What is the formula for finding the area of a curve?

The formula for finding the area of a curve is known as the integral. It is represented as ∫f(x)dx, where f(x) is the function being integrated and dx represents the infinitesimal change in the independent variable, x.

2. How do you find the limits of integration for an integral?

The limits of integration for an integral are determined by the boundaries of the region being integrated. These boundaries can be given explicitly or can be inferred from the graph of the function.

3. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and gives a single numerical value as the result. An indefinite integral does not have limits of integration and represents a family of functions rather than a single value.

4. Can you use the fundamental theorem of calculus to evaluate an integral?

Yes, the fundamental theorem of calculus states that the definite integral of a function can be evaluated by finding its antiderivative and evaluating it at the limits of integration. This allows for the evaluation of difficult integrals by using simpler antiderivatives.

5. How can you check if your answer for an integral is correct?

One way to check if your answer for an integral is correct is by using the properties of integrals. For example, you can check if your answer is reasonable by comparing it to the graph of the function or by using the symmetry property. You can also take the derivative of your answer and see if it matches the original function.

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