Substitution method for finding an integral's interval changes

[thegreengineer]

Look, I was wondering if substituting the variable more than once is valid and hence the definite integral intervals change this way.
Consider the following integral (I'm working for finding the volume of a solid of revolution):
*\pi \int_{-3}^{5}3^{2}-(\sqrt{\frac{y+3}{2}}+1)^2dy
Personally I hate to work hard (in this case I mean I would not rather to expand that binomial (\sqrt{\frac{y+3}{2}}+1)^2 because it has a division within a square root and we also have a binomial which is y+3 so it would be so much time-wasting in my opinion to expand all that. So what I did was to transform that y+3 into u so:
u=y+3
du=dy
Also I changed the intervals (if u=y+3 then the new interval would be obtained by substituting the values within the y+3 so the first interval would be a=(-3)+3=0 and the other b=5+3=8 so the integral would become:
*\pi \int_{0}^{8}3^{2}-(\sqrt{\frac{u}{2}}+1)^2du
This integral now it seems easier than the original, however, it's still kind of time wasting in my opinion to have a division within the root, so now I convert \frac{u}{2} into w so:
w=\frac{u}{2}
dw=\frac{du}{2}
Since I made another substitution then the integral interval must be changed once again so if w=\frac{u}{2} then the first limit of the interval is got by replacing the value of the first limit of the last integral into w=\frac{u}{2} so a=\frac{0}{2}=0 and b=\frac{8}{2}=4, therefore the integral becomes:
*(1/2)\pi \int_{0}^{4}3^{2}-(\sqrt{w}+1)^2dw.

I don't know if this is valid, I would appreciate if someone told me because I'm failing in integral calculus. Thanks.

 

[Mark44]

Look, I was wondering if substituting the variable more than once is valid and hence the definite integral intervals change this way.
Consider the following integral (I'm working for finding the volume of a solid of revolution):
*\pi \int_{-3}^{5}3^{2}-(\sqrt{\frac{y+3}{2}}+1)^2dy
Personally I hate to work hard (in this case I mean I would not rather to expand that binomial (\sqrt{\frac{y+3}{2}}+1)^2 because it has a division within a square root and we also have a binomial which is y+3 so it would be so much time-wasting in my opinion to expand all that. So what I did was to transform that y+3 into u so:
u=y+3
du=dy
Also I changed the intervals (if u=y+3 then the new interval would be obtained by substituting the values within the y+3 so the first interval would be a=(-3)+3=0 and the other b=5+3=8 so the integral would become:
*\pi \int_{0}^{8}3^{2}-(\sqrt{\frac{u}{2}}+1)^2du
This integral now it seems easier than the original, however, it's still kind of time wasting in my opinion to have a division within the root, so now I convert \frac{u}{2} into w so:
w=\frac{u}{2}
dw=\frac{du}{2}
Since I made another substitution then the integral interval must be changed once again so if w=\frac{u}{2} then the first limit of the interval is got by replacing the value of the first limit of the last integral into w=\frac{u}{2} so a=\frac{0}{2}=0 and b=\frac{8}{2}=4, therefore the integral becomes:
*(1/2)\pi \int_{0}^{4}3^{2}-(\sqrt{w}+1)^2dw.

I don't know if this is valid, I would appreciate if someone told me because I'm failing in integral calculus. Thanks.

Yes, it's valid to make more than one substitution, changing the limits of integration each time.

It would be simpler, though, to just make one substitution: $w = \frac{y + 3} 2$. Then dw = dy/2.

Alternatively, expanding $(\frac{y + 3}{2} + 1)^2$ isn't really that hard. You probably could have done it in the time it took you to type your question. After expanding, get common denominators (of 4), and bring out a factor of 1/2 outside the radical.

 

[wabbit]

(Previous comment deleted, just saw Mark44 said prerty much the same)

There is just one thing in your last formula, you got a factor wrong - note that $dw=\frac{du}{2}$ so $du=2dw$.