Question

Calculus 3!!!! Help

1. Compute the line integral of the vector field
F= < y2+2-cox(x4) , 3+sin (y6) +4xy> , for the closed curve consisting of the x-axis from ( 0 , 0 ) to ( 3 , 0 ) , followed by the quarter circle (centered at the origin) from ( 3 , 0 ) to ( 0 , 3 ), and back to ( 0 , 0 ) along the y-axis.


2. For the solid S bounded by x= y2+ z2 and x = 9 , find: a) the volume; b) the centroid.

Answer 1

wheres the vec field

Answer 2

Compute the line integral of the vector field
F= < y2+2-cox(x4) , 3+sin (y6) +4xy> , for the closed curve consisting of the x-axis from ( 0 , 0 ) to ( 3 , 0 ) , followed by the quarter circle (centered at the origin) from ( 3 , 0 ) to ( 0 , 3 ), and back to ( 0 , 0 ) along the y-axis.


For the solid S bounded by x= y2+ z2 and x = 9 , find: a) the volume; b) the centroid.

Answer 3

The Question did not copy over the first time sorry

Answer 4

parametrize the curve ur integrating over

Answer 5

we will get it in this form
F(x(t),y(t)) dot r'(t)/|r'(t)| dt

Answer 6

r(t) is our parametrized curve and we will rewrite our vector field F in terms of our parameter

Answer 7

F.dR

dR=dR/dt * dt

Answer 8

go ahead and parametrize, or find the bounds
a circle with radius 1
x=cost , y= sin t
a horizontal line on xaxis is of form
x=t,y=0
use this to figure out your curve

Answer 9

Okay. I will try that

Answer 10

Green's theorem, if you have covered it already, works nicely here..

Answer 11

Green's seems to be the only choice. Integrating along the horizontal part would be difficult without resorting to numerical methods:
\[\int_{C_1}\mathbf F\cdot\mathrm d\vec{r}_1=\int_0^1\left\langle2-\cos(81 t^4),3\right\rangle\cdot\left\langle3,0\right\rangle\,\mathrm dt=3\int_0^1(2-\cos(81t^4))\,\mathrm dt\]There is an antiderivative in terms of the incomplete gamma function, but I doubt that would be covered in a Calc III course.
http://mathworld.wolfram.com/IncompleteGammaFunction.html
If you write \(\mathbf F(x,y)=\langle P(x,y),Q(x,y)\rangle\), then by Green's theorem you have
\[\int_C\mathbf F\cdot\mathrm d\vec{r}=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,\mathrm dx\,\mathrm dy\]where \(D\) is the region bounded by \(C\).

With \(\mathbf F=\left\langle \underbrace{y^2+2-\cos x^4}_P,\underbrace{3+\sin y^6+4xy}_Q\right\rangle\), you have
\[\int_C\mathbf F\cdot\mathrm d\vec{r}=\iint_D(4y-2y)\,\mathrm dx\,\mathrm dy=2\iint_Dy\,\mathrm dx\,\mathrm dy\]which can be computed easily if you convert to polar coordinates.