Question

show work done...... question in picture at second posting ty.....

Answer 1

@ash2326

Answer 2

We have
\[F=zi+xj-yk\]
\[r=(\cos t) i+(\sin t)j+ (t)k\]
\[dr= [(-\sin t) i+(\cos t)j+(1)k]dt\]
now
\[ F(r)=F(\cos t, \sin t , t)= (t)i+(\cos t) j+ (-\sin t) k\]
\[ F(r).dr=((t)i+(\cos t) j+ (-\sin t) k).((-\sin t) i+(\cos t)j+(1)k)dt\]
We get
\[ F(r).dr= (-t \sin t+\cos^2 t-\sin t)dt\]
\[ \int F.dr= \int_{0}^{2\pi}(-t \sin t+\cos^2 t-\sin t)dt\]
Let's evaluate the integral
\[\int_{0}^{2\pi}(-t \sin t+\cos^2 t-\sin t)dt\]
We know that
\[\int_{0}^{2\pi}(-\sin t)dt=0\]
\[I=\int_{0}^{2\pi}(\cos^2 t)dt\]
Using definite integral property
\[If \int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x)=2\int_{0}^{a/2} f(x) dx\]
Using this twice we get
\[I=\int_{0}^{2\pi}(\cos^2 (t))dt=4\int_{0}^{\pi/2}(\cos^2 (t))dt\]
\[I=4\int_{0}^{\pi/2}(\sin^2 (t))dt\]
If we add the two, we'll get
\[2I=4\int_{0}^{\pi/2}(\sin^2 (t)+\cos^2 (t))dt=4\times \pi/2=2\pi\]
We get
\[I= \pi]\]
Also the integral
\[\int_{0}^{2\pi} -t \sin t dt=2\pi\]
So we
get
\[\int F. dr= 2\pi+\pi+0=3\pi\]

Answer 3

when comes to the F(r) part how u get (t)i+(cost)j+(−sint)k do clarify ty...:)

Answer 4

\[r=\cos t i+ \sin t j+ t k=(\cos t, \sin t, t)\]
or
\[x= \cos t , y= \sin t , z= t\]
substitute these in F to find F(r)
\[F= zi+ xj-yk= t i+ \cos t j- \sin t k\]

Answer 5

oh i see the x y and z are not in order heiz haha my mistake kk ...continuing

Answer 6

ok done i understand till the end wow you explain clearly and step by step great :)

Answer 7

welcome :D