Look, I was wondering if substituting the variable more than once is valid and hence the definite integral intervals change this way.(adsbygoogle = window.adsbygoogle || []).push({});

Consider the following integral (I'm working for finding the volume of a solid of revolution):

*[itex]\pi \int_{-3}^{5}3^{2}-(\sqrt{\frac{y+3}{2}}+1)^2dy[/itex]

Personally I hate to work hard (in this case I mean I would not rather to expand that binomial [itex](\sqrt{\frac{y+3}{2}}+1)^2[/itex] because it has a division within a square root and we also have a binomial which is [itex]y+3[/itex] so it would be so much time-wasting in my opinion to expand all that. So what I did was to transform that [itex]y+3[/itex] into [itex]u[/itex] so:

[itex]u=y+3[/itex]

[itex]du=dy[/itex]

Also I changed the intervals (if [itex]u=y+3[/itex] then the new interval would be obtained by substituting the values within the [itex]y+3[/itex] so the first interval would be [itex]a=(-3)+3=0[/itex] and the other [itex]b=5+3=8[/itex] so the integral would become:

*[itex]\pi \int_{0}^{8}3^{2}-(\sqrt{\frac{u}{2}}+1)^2du[/itex]

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 [itex]\frac{u}{2}[/itex] into [itex]w[/itex] so:

[itex]w=\frac{u}{2}[/itex]

[itex]dw=\frac{du}{2}[/itex]

Since I made another substitution then the integral interval must be changed once again so if [itex]w=\frac{u}{2}[/itex] then the first limit of the interval is got by replacing the value of the first limit of the last integral into [itex]w=\frac{u}{2}[/itex] so [itex]a=\frac{0}{2}=0[/itex] and [itex]b=\frac{8}{2}=4[/itex], therefore the integral becomes:

*[itex](1/2)\pi \int_{0}^{4}3^{2}-(\sqrt{w}+1)^2dw[/itex].

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

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# Substitution method for finding an integral's interval changes

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