Recent content by helpm3pl3ase

hm what I came up with, which is probably wrong, is almost like a triangle in the first quad... so therefore you would go from (pie/2)/2 = pie/4 to 0??

iam not so sure i know how to sketch it.

What if i tried not to use polar and did this.. from above we have: ∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx ∫(2 to 1) 1/3[(x^2 + y^2)^(3/2)] |(y=x and y=0) dx 1/3 ∫(2 to 1) x^3 dx 1/3 x 1/4 x^4 | (x = 2 and x =1) final answer: 5/4

yes it worked, got 5 and 15

oh its a circle that would go from 2pie to 0.. but iam still confused about the other part

I got the (r^3)^(1/2) from multiplying the r on the outside to the inside.. the (1/2) is the sqrt and the r^3 is from multiplying it with the r.. also the theta limit would be r to 0, rather than x to 0??

<3, -6, -2> X <2, 1, -2> = <14, 2, 15> Then set z to 0 to get x = 3, y = -1 ==> <3, -1, 0> x = 3 + 14t y = -1 + 2t z = 15t

so ∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx ∫(2 to 1)∫(x to 0) ((r sin(theta)(r^2)^1/2)r dr dtheta ∫(2 to 1)∫(x to 0) ((r^2 sin(theta)(r^3)^1/2) dr dtheta Like this?

Homework Statement Evaluate the integral ∫∫R y√(x^2 + y^2) dA with R the region {(x, y) : 1 ≤ x^2 + y^2 ≤ 2, 0 ≤ y ≤ x.} Homework Equations The Attempt at a Solution Solution: ∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx ∫(2 to 1) 1/3[(x^2 + y^2)^(3/2)] |(y=x and y=0)...

Homework Statement Find parametric equations for the line in which the planes 3x − 6y − 2z = 15 and 2x + y − 2z = 5 intersect. Homework Equations The Attempt at a Solution <2, 1, -2> - <3, -6, -2> = <-1, 7, 0> x = 2 - t, y = 1 + 7t, z = -2 Did I do this correctly??

Hey hows it going?? I am having some trouble on this problem: Use Green's Theorem to evaluate the line integral ∫C F . dr where F =< y^3 + sin 2x, 2x(y^2) + cos y > and C is the unit circle x^2 + y^2 = 1 which is oriented counterclockwise. I started like so: ∫C Pdx + Qdy = ∫∫D Qx...

ahhhhhhhhhhhhhh alright i think i got it.. thank you all for your help.. But what happens to the 7?

i get how you got 7[3 0 -4] but for the answer it shows: (1/5) [3 0 -4] So I am still not sure how the hell they got the (1/5)

I did this.. just want to make sure my formula is right for vector 2.. where u1 = what you received for the first vector. V2 = v2 - (u1 (dot product) v2)u1 ~~~~~~~~~~~~~~~~~~~~ ||v2 - (u1 (dot product) v2)u1|| where ~~~~ = divide. I did this over and over again and seem to get (1/35)v2

blahh.. I don't get it.. I did and still not the right answer is produced.. can someone help.. I don't know what I am missing The main question is to just perform Gram-Schmidt on those first two vectors: 4 0 3 and 25 0 -25 I did it in the first post and just did it again.. I still i get a...