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Let $D=\left\{(y,z): \sqrt{y} \leq z \leq 2-\sqrt{y} \right\}$. Rotate $D$ $360$ degree around the $z$ axis generating the volume $V_z$ and evaluate the lateral surface of $V_z$, $|\partial V_z|$.

For evaluating the lateral surface of $V_z$, I considered these two curves:

  • $\gamma_1(t) = (t,\sqrt{t})$, $0 \leq t \leq 1$;
  • $\gamma_2(t) = (t,2- \sqrt{t})$, $0 \leq t \leq 1$.

So I continued by doing the integral $$|\partial V_z| = 2\pi\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}dt +2\pi\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}dt = 4\pi\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}dt.$$

Any tips on how to solve this integral? Is it possible to evaluate the surface with a different approach?

migliio
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  • You can get it to standard $\int \sqrt{x^2-a^2}$ form and then either apply the formula or integrate by parts. – Math Lover Feb 01 '21 at 21:39

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The surface area of a solid of revolution around the $z$-axis is given by $$S = 2\pi\int_{a}^{b} f(z) \sqrt{1 + f'(z)^{2}}\, dz.$$ In your case we have two functions: $f_1(z)=z^2$ (from $z=\sqrt{y}$) over the interval $[0,1]$ and $f_2(z)=(2-z)^2$ (from $z=2-\sqrt{y}$) over the interval $[1,2]$. Therefore $$|\partial V_z|=2\pi\int_{0}^{1} f_1(z) \sqrt{1 + f_1'(z)^{2}}\, dz+2\pi\int_{1}^{2} f_2(z) \sqrt{1 + f_2'(z)^{2}}\, dz$$ Note that, by symmetry, the two integrals are equal. Moreover $$\int_{0}^{1} f_1(z) \sqrt{1 + f_1'(z)^{2}}\, dz=\int_{0}^{1} z^2 \sqrt{1 + 4z^2}\, dz=\int_{0}^{\text{arcsinh}(2)}\cosh^2(t)\sinh^2(t)\,dt$$ where we applied the substitution $2z=\sinh(t)$. Can you take it from here? You may also take a look at How to evaluate integral: $\int x^2\sqrt{x^2+1}\;dx.$

P.S. Your approach is correct and after letting $t=z^2$ we find the same integral $$\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}\,dt=\int_{0}^{1} z^2 \sqrt{1 + 4z^2}\, dz.$$ I don't see an easier way to find this area.

Robert Z
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  • Thanks. Starting from your hint, $$I = \int_{0}^{\log(1+\sqrt{2})}\frac{1}{8} \left( \cosh(4t)-1 \right)dt = \dots = \frac{1}{32}\left[ \sinh(4t)\right]_{0}^{\log(1+\sqrt{2})}-\frac{1}{8}\left(\log(1+\sqrt{2}) \right ) = \frac{3\sqrt{2}-\log(1+\sqrt{2})}{8}.$$ So that $$|\partial V_z|=4\pi \cdot I = \pi\frac{3\sqrt{2}-\log(1+\sqrt{2})}{2}.$$ Is that right? – migliio Feb 02 '21 at 14:11
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    Yes, as soon as we take the correct upper limit $\text{arcsinh}(2)=\log(\sqrt{5}+2)$ (I edited my answer with the right one, sorry) – Robert Z Feb 02 '21 at 14:40