Question

Find the volume of the region bounded by the paraboloid z=X^(2)+y^(2) and below the triangle enclosed by the lines y=x,x=0,and x+y=2 in the xy-plane.

Answer 1

@greatlife44 have time to help me with this?

Answer 2

how much calculus do you know @Johan14th ?

Answer 3

Sorry not really sure on this one :(

Answer 4

need to know multi-variable calculus to do this. That's where you learn how to do this problem. no worries

Answer 5

This takes me back ^^
\[\Large \iint_R x^2+y^2dA\]

Answer 6

@Ac3
Still here, are you? ^^

Answer 7

Well, while you're still somewhere else, I might as well sketch a graph of the region R in question...

Answer 8

ok good i set up the integral as....

Answer 9

\[\int\limits_{0}^{1}\int\limits_{x}^{2-x}x ^{2}+y ^{2}dydx\]

Answer 10

and when i solved i got (23/18) which is wrong answer is (4/3)

Answer 11

Hang on, let's see:
\[\Large = \int_0^1\left[x^2y + \frac13y^3\right]_{y=x}^{y=2-x}\]

Answer 12

(2-x)^3 expanded = 8-4x-8x+4x^(2)+2x^(2)-x^(3) you can double check me though

Answer 13

unless i set up the integral incorrectly

Answer 14

dx.

Oops.

Anyway, this becomes
\[\large = \int_0^1\left[x^2(2-x) + \frac13(2-x)^3\right]-\left[x^3 + \frac13x^3\right]dx\]

Answer 15

ok nice that's what i had

Answer 16

I'm sorry ^^
I'm not usually this slow, but I haven't done this in a long while :D

Answer 17

no worries man so far so good you have exactly what i got only i got it wrong lol

Answer 18

i'm not sure if i set up that double integral properly

Answer 19

Means I probably will too :(

Oh well

\[\large = \int_0^1\left[2x^2-x^3 + \frac13(8 -12x+6x^2-x^3)\right]-\left[x^3 + \frac13x^3\right]dx\]

Answer 20

\[\large = \int_0^1\left[2x^2-x^3 + \frac83 -4x+2x^2-\frac{x^3}3\right]-\left[x^3 + \frac{x^3}3\right]dx\]

Answer 21

hmm keep going maybe you'll get the right answer

Answer 22

Judging by that reaction, have I deviated from your solution?

Answer 23

so far yes

Answer 24

maybe i calculated it incorrectly

Answer 25

In that case
\[\Large \int_0^1 \left[-\frac83 x^3 + 4 x^2-4x+
\frac83\right]dx\]

Answer 26

YOU GOT IT!!!

Answer 27

ah dam it so i did set it up right i just have to make sure i simplify properly

Answer 28

Should be \[\Large \left[-\frac23x^4 +
\frac43x^3-2x^2 +
\frac83x\right]_0^1\]

Answer 29

well done

Answer 30

Looks like I'm not that rusty after all ^^
It has been years, you know XD

Answer 31

how do i give you a medal

Answer 32

you just did ^^
Well... this was fun :D

Answer 33

Don't you just HATE these problems which are so susceptible to human error? :D

Answer 34

oh so that's how you do it. Yes i hate them now i'm gonna re-do and make sure i get the right answer

Answer 35

it's like they hand-craft them on purpose

Answer 36

Well, see ya around ^^