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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

I start with \(\displaystyle x=\sqrt{28y}\) then \(\displaystyle f'(x)=\frac{14}{\sqrt{28y}}\)

so we got

\(\displaystyle 2\pi\int_0^{21}\sqrt{28y}\sqrt{1+(\frac{14}{\sqrt{28y}})}^2\)

then I rewrite as \(\displaystyle \int_0^2\sqrt{28y}\sqrt{1+\frac{196}{28y}}\)

\(\displaystyle \sqrt{28y}\sqrt{1+\frac{196}{28y}}<=>\sqrt{28y+196}\)

So I got \(\displaystyle 2\pi\int_0^{21}\sqrt{28y+196}\) and if Integrate it I get

\(\displaystyle \frac{(28y+196)^{1.5}}{1.5*28}\) Is this correct?