Help with Integral: Find the area of C

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so have you tried anything?
 
here's how you can write it in tex
\int_C y dx-x dy +dz <br />

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

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

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

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

So the integral becomes \int_C [b*sint*dx-a*cost*dy+dz]
Which equals \int_C -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?
 
lanedance said:
here's how you can write it in tex
\int_C y dx-x dy +dz <br />

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

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

Or, a little neater write the integrand

\langle y, -x, 1\rangle \cdot \langle dx,dy,dz\rangle
 
gipc said:
So the integral becomes \int_C [b*sint*dx-a*cost*dy+dz]
Which equals \int_C -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?
 
gipc said:
So the integral becomes \int_C [b*sint*dx-a*cost*dy+dz]
Yes, that is correct.

Which equals \int_C -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?
 
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