Parametrizing a Cylinder: Solving for the Part Between Two Planes

[hils0005]

Homework Statement

Parametrize the part of the cylinder 4y^2 + z^2 = 36 between the planes x= -3 and x=7

The Attempt at a Solution

radius=6
Parametric equations:
x=x
y=4 + 6cos(theta)
z=6sin(theta)

in vector form
\widehat{}r= x\widehat{}i + (4 + 6cos(theta))\widehat{}j + 6sin(theta)\widehat{}k

-3 \leq x \leq 7

I really don't know if I'm completing correctly, any direction would be appreciated. Thanks!

 

[Pere Callahan]

Homework Statement

Parametrize the part of the cylinder 4y^2 + z^2 = 36 between the planes x= -3 and x=7

The Attempt at a Solution

radius=6
Parametric equations:
x=x
y=4 + 6cos(theta)
z=6sin(theta)

These y,z do NOT satisfy 4y^2+z^2=36. Moreover, you cannot speak of a radius here, as the crosssection of the cylinder is not a circle.

 

[hils0005]

the cross sections of the cylinder perpendicular to the x-axis are circles y^2 + z^2=6^2 correct?
then this would be given parametrically as y=6cos(theta) and z=6sin(theta) right?

 

[Pere Callahan]

This is correct, but in your original problem the y^2 is multiplied by a factor 4.
If you let y'=2y, then you have

y'^2+z^2=36 and so

y'=6 cos(theta)
z =6 sin(theta)

Now use y=y'/2 and you're done.

 

[hils0005]

Thanks I did not know what to do with factor of 4

so would this be correct?
x=x
y=3cos(theta)
z=6sin(theta)

in vector form
\widehat{}r= x\widehat{}i + 3cos(theta)\widehat{}j + 6sin(theta)\widehat{}k

-3 \leq x \leq 7

do i need to say anything about theta going from 0 to 2pi?