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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):

*[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.

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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