Question

Find the volume of the region under the graph of f(x,y) = 5x+y+1 and above the region y^2\le x, 0\le x\le 4.

Answer 1

@myininaya the second part is \[y^2 \le x, 0 \le x \le 4\]

Answer 2

myninia i took ap statistics and they kicked me out cause statitics test is in may

Answer 3

so i think id set it up as\[\int\limits_{0}^{4} \int\limits_{?}^{?} 5x+y+1 dy dx\]

Answer 4

i just dont know what the y bounds would be if that is correct. one might be y=x^2

Answer 5

The region \(y^2\leq x, 0\leq x\leq 4\)?

Answer 6

That is a weird region

Answer 7

i know

Answer 8

I think I would use \(y\leq \sqrt x\).

Answer 9

The top limit will give \(y=-5x-1\).

Answer 10

It doesn't make sense, because the top part is below the bottom part.

Answer 11

Hmm, wait up, this is volume so it is a triple integral...

Answer 12

double integral

Answer 13

Maybe use:\[\int\limits_{0}^{4} \int\limits_{0}^{\sqrt x} 5x+y+1~ dy dx\]

I don't like how they say "above the region". I think they mean within the region.
I also don't like how there is no seeming lower limit for \(y\).

Answer 14

naa that didnt work

Answer 15

On other thing you could try then:\[\int\limits_{0}^{4} \int\limits_{-\sqrt x}^{\sqrt x} 5x+y+1~ dy dx\]

Answer 16

http://www.wolframalpha.com/input/?i=%5Cint+%5Cint+5x%2By%2B1+dy+dx%2C+y+%3D+-sqrt%28x%29+to+sqrt%28x%29%2C+x+%3D+0+to+4

Answer 17

swag

Answer 18

did that work @pmkat14 ?
just curious myself

Answer 19

yea!!!!