Hey so I am having trouble with questions 1, 3, and 6.



For question 1 I've already showed that you can solve for u and v as functions of x and y. Let the first function be f and the second function be g. The matrix composed of partials of f and g with respect to u and v (dunno how to format it here so just describing with words) is invertible so I've got that part done. What I need help with is computing all partial derivatives of u and v at (1,1). I think I have to implicitly differentiate f and g and then end up with some equations where I can find u' and v' but I'm not sure how to get there.

For 3 I tried drawing it out at first but then I found saw that the tetrahedron formed has no symmetries that make it easy to work with. I think I may have to use a couple of double integrals to find the volume but I'm having trouble figuring out the bounds of integration. I'm not even sure what function I would use for the integrand, would I use the equations of the planes formed by the faces of the tetrahedron?

For question 6 x=0 should actually be y=0
And for 6 I'm having trouble figuring out the shape. I think with z = 5-x^2 and z = 4x^2 I get some sort of tube shape going along the y-axis, and then it is chopped with the x+y=1 plane. I think I would use the region bounded by z=5-x^2 and z=4x^2 as my "base" and then have the x+y=1 plane as the height of the cylinder thing. Not really sure though

Can anyone provide some insight?

Thanks!!

Actually solved #6, got volume = 20/3. Still need help with 1 and 3

3. The volume of your tetrahedron is given by the formulae : V = Base*Heigth /3, just like any other pyramid.
Considering the base on the (xz) plane :
Base = 3*4 /2 = 6
Heigth =1

so volume = 6*1 / 3 = 2

The use of any double integral is simply useless in this case.

6. It seems to me that your volume is not bounded properly. Its volume is infinite.

This post was edited by feanur on Nov 27 2014 07:17pm

For question 6 x=0 should actually be y=0.

Well in this case :

Volume = Integral ( y=0 to 1, x = 0 to 1-y, z = 4x² to 5-x², dydxdz)

V = I ( y=0 to 1, x=0 to 1-y, (5-5x²)dxdy )

V = I ( y=0 to 1, [5x - 5x^3/3] between 0 and 1-y.dy )

V = I ( y=0 to 1, [5(1-y) - 5(1-y)^3/3].dy )

3V = I (y=0 to 1, (10 - 15y² + 5y^3)dy )

3V = [ 10y - 5y^3 + (5/4)y^4 ] between 0 and 1

3V = 10 - 5 + (5/4) = 25/4

V = 25/12

if I do no mistake.

I considered the region bounded between z = 4x² and z = 5-x² as the base of the solid and the top piece was the plane x+y=1. So

V = I (x=-1 to 1, z=4x² to 5-x², y=0 to 1-x) dydzdx

Where do you get y=0 to 1 from?

not sure how you have x=-1 to 1 because x is bounded by the plane y=0
feanur's limits seem to be correct

This post was edited by rwarth on Nov 27 2014 09:09pm

are you saying the points (-1,0,0) and (1,0,0) do not lie on the plane y = 0?

those points do lie on the plane y=0.
the plane y=0 is constructed by taking the line y=0 and extending it across x and z.
however you will be crossing the plane when you go from (-1,0,0) to (1,0,0) and therefore they both wouldn't be in your boundary