Question

Set up a double integral that corresponds to the mass of the semi disk

(give me a min to post the formula)

Answer 1

\[D=(x,y) : 0 \le y \le \sqrt{4-x ^{2}}\]

if the density at each of its points is given by the function \[\delta (x,y)= x ^{2}y\]

Answer 2

\[\int\limits_{-2}^{2} \int\limits_{0}^{\sqrt{4-x ^{2}}} x ^{2}y* dy dx\]

this is the answer

Answer 3

@hartnn

Answer 4

anybodyyyyyyy

Answer 5

Well this one I kinda get, mass is the integral of density over volume.

Answer 6

The y limits come from this \[\Large 0 \le y \le \sqrt{4-x ^{2}}\]so that gives \[\Large \int\limits\limits_{0}^{\sqrt{4-x ^{2}}} x ^{2}y* dy\](integrating the density over the y limits)

Answer 7

the next question is find the mass of the above semidisk

Answer 8

yeah but how did they know where the endpoint to the integral go for a and b?

Answer 9

And the x-limits come from the \[\Large \sqrt{4-x ^{2}} \]since you can only get real numbers for y if -2<x<2 right?

Answer 10

yeah but is that for every case?

Answer 11

Every case...?

Answer 12

and how do u find the mass now?

Answer 13

every time i get these kinds of problems?

Answer 14

In this case it's because of that sqrt(4-x^2)... it's only real numbers if x is between -2 and 2, since square roots of negative numbers are imaginary.

Okay the mass means we need to integrate this \[\Large \int\limits\limits_{-2}^{2} \int\limits\limits_{0}^{\sqrt{4-x ^{2}}} x ^{2}y* dy dx\]

Answer 15

It might make it easier to change the order of integration (you can do that with double integrals), do dx first then dy.
\[\Large \int\limits\limits_{0}^{\sqrt{4-x ^{2}}}\int\limits\limits_{-2}^{2} x ^{2}y* dx dy \]

Answer 16

you know you can't do that ^, right ?

Answer 17

Wait... you can't? I thought it was valid...

Answer 18

i mean you can't change the order like that..

Answer 19

and why do u even need to change the order ?

Answer 20

its a really easy double integral to solve,
integrate w.r.t y first, treating x as constant

Answer 21

i took the integral from 0 to the sqrt 4-x^2 and integrated it which came to x^3*y/3 and then dont i just solve it from 0 to the sqrt?

Answer 22

and then once i get that answer integrate it from -2 to 2?

Answer 23

\[\Large \int\limits\limits_{0}^{\sqrt{4-x ^{2}}} x ^{2}y* dy = \left[ \frac{ 1 }{ 2}x^2 y^2 \right]_{0}^{\sqrt{4-x^2}}\]oh right i was thinking the square root makes things awkward but it doesn't

Answer 24

Meg you integrated wrt x not y

Answer 25

whats the difference? ive nvr been taught that...well tht i know of

Answer 26

\[\Large \left[ \frac{ 1 }{ 2}x^2 y^2 \right]_{0}^{\sqrt{4-x^2}} = \frac{ 1 }{ 2}x^2(4-x^2)\]

Answer 27

wait idk how to integrate that

Answer 28

dy means you integrate y, not x. Like y integrates to y^2/2, same way x integrates to x^2/2

Answer 29

well x is from -2 to 2, and y is from 0 to sqrt(4-x^2), you have the limit of integration.

Answer 30

how did u come up with that integration?

Answer 31

Treat x as a constant. increase power on y, divide by new power\[\Large \int\limits\limits\limits_{0}^{\sqrt{4-x ^{2}}} x ^{2}y* dy = \left[ \frac{ 1 }{ 2}x^2 y^2 \right]_{0}^{\sqrt{4-x^2}}\]

Answer 32

still dont get it

Answer 33

:/ sorry

Answer 34

oh wait i see it