Question

how to solve the integration of y*(9-y^2)^1/2

Answer 1

put y^2=t

Answer 2

there is a sqrt on (9-y^2) do i have to solve this question with integeration by parts??

Answer 3

not really.
if you put y^2=t
you have ydy=dt/2
and hence you will have to integrate ((9-t)^1/2)/2
which looks easier to solve

Answer 4

this was the question of multiple integrals, i have solved one integral and i get this, the anx is 9 bt i am not able to solve it

Answer 5

the limits on the right integral are functions of y. That means the limits shows the range of x. therefore integrate at first with dx and then with dy.
you should get 9 as an answer finally.

Answer 6

if i put y^2=t i will also have to change the limits

Answer 7

yes i integrated it first with dx

Answer 8

o to (9-y^2)^1/2 is the range of x

Answer 9

one you are done integrating with x, you can put y^2=t and that will not change limit. coz limits are constants.

Answer 10

*once

Answer 11

you may try "9-y^2 = t" also

Answer 12

yeah, that is more simpler.

Answer 13

ok i'll try

Answer 14

i am not getting the correct anx by putting y^2=t and the same limits

Answer 15

what are you getting?

Answer 16

the anx is in cube root or if i simplify it it is in points -0.087....

Answer 17

after putting 9-y^2 =t
find dt at first and tell me what you get

Answer 18

ok

Answer 19

-cuberoot 9/3

Answer 20

just cuberoot 9 /3

Answer 21

find dt in terms of dy at first

Answer 22

dt/dy=-2y i have done the whole method and the limits are not constant, because it is iterated integration

Answer 23

the limits after first integration is constant.
put 9-y^2=t
so, now you need to integrate (sqrt t)/-2 in terms of dt

Answer 24

put 9-y^2 =t after you are done integrating with x

Answer 25

\[\int_{0}^{3}\int_{0}^{\sqrt{9-y^2}}ydxdy\]
\[=\int_{0}^{3}y\sqrt{9-y^2}dy\]
Now put \(9-y^2=t\). Therefore \[ydy=-\frac{dt}{2}\]
So, now you need to integrate
\[\int_{0}^{3}\frac{\sqrt{t}}{-2}dt\]
You can integrate this, right?

Answer 26

yes i did the same the anx is not 9

Answer 27

oh, sorry, the limits change.
when you put 9-y^2=t,
For y=0, you have t=9
and when y=3 then t=0
So, the limit is 9 to 0
you need to integrate \[\int_{9}^{0}\frac{\sqrt{t}}{-2}dt\]
you should get 9 now.

Answer 28

or else you can switch back to x later on after integration, i.e express t in terms of x again and keep the limits unchanged.