Integral question

Question

Integral question

hba · Answer

$$\int\limits_{0}^{3} \int\limits_{0}^{1} x(x^2 + y)^{1/2} dxdy$$

sidsiddhartha · Answer

it means
x=0,x=3
and
y=0,y=3

hba · Answer

u=x^2+y

hba · Answer

Substitution right?

dumbcow · Answer

For first integral, treat "y" as a constant

yes that substitution should work

sidsiddhartha · Answer

just separate them and integrate , i dont think substitution is needed

hba · Answer

Thankyou :)

sidsiddhartha · Answer

$$\int\limits_{0}^{3}[\int\limits_{0}^{1}x(x^2+y)^{1/2}.dy].dx$$

hba · Answer

When i integrated it with respect to x and applied the limits i got [(1+y)^ 3/2 ]/3

hba · Answer

Is it correct?

dumbcow · Answer

what about the other limit when x=0

sidsiddhartha · Answer

yes u need the change the limits

hba · Answer

$$\frac{ (1+y )^{3/2}} {3  } -y$$

dumbcow · Answer

$$\frac{(1+y)^{3/2} - y^{3/2}}{3}$$

hba · Answer

Yeah i got this sorry

hba · Answer

Should i use substitution again when i do it with y now?

dumbcow · Answer

yes for the "1+y" part

u = 1+y
du = dy

hba · Answer

Yeah i am using the same

hba · Answer

But how will i cater y^3/2?

hba · Answer

u+1?

dumbcow · Answer

treat it as a separate integral

hba · Answer

Oh okay okay

hba · Answer

The answer turns out to be so complicated i feel it's wrong

dumbcow · Answer

do you get
$$\frac{2}{15} (4^{5/2} - 3^{5/2} - 1)$$

hba · Answer

i got something like 64/15 - 18 root(3) /15

dumbcow · Answer

yeah same thing except you dont have the "-1" 

remember when doing the limits you have to apply the 2nd limit of y=0

hba · Answer

okay i got it thankyou :)

dumbcow · Answer