Find the surface area cut from the cone$\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{5}{\mathbf{z}}^{\mathbf{2}}\mathbf{,}\mathbf{z}\mathbf{>}\mathbf{0}$by the cylinder${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{y}$

The equation of cone is $2x^2+2y^2=5z^2,z>0$and the equation of cylinder is$x^2+y^2=2y$ .

An intersection curve consists of the common points of two transversally intersecting surfaces. The surface area of a three-dimensional object is the total area of all its faces.

The equation of the circle is$x^2+{(y-1)}^2=1$

The equation of cone is$\varphi =5z^2-2x^2-2y^2$

The angle is calculated as follows:

$$\begin{gathered} sec\gamma =\frac{\sqrt{{\left|\nabla \varphi \right|}^2}}{\left|\partial \varphi /\partial z\right|} \\ =\frac{\sqrt{\left|10z\hat{z}-4x\hat{x}-4y\hat{y}\right|}}{10z} \\ =\frac{\sqrt{100z^2+16x^2+16y^2}}{10z} \end{gathered}$$

From the equation of the cone, we have the following.

$$\begin{gathered} z^2=\frac{2}{5}{r}^2 \\ \Rightarrow r^2=\frac{5}{2}{z}^2 \end{gathered}$$

Substitute this back into angle equation.

$$\begin{gathered} sec\gamma =\frac{\sqrt{100z^2+16x^2+16y^2}}{10z} \\ =\frac{\sqrt{100z^2+16r^2}}{10z} \\ =\frac{\sqrt{100+40}}{10} \\ =\sqrt{\frac{7}{5}} \end{gathered}$$

The required area is calculated as below.

$$\begin{gathered} A={\int }_0^{X}d\theta {\int }_0^{2\sin \theta }rdrsec\gamma \\ =\sqrt{\frac{7}{5}}{\int }_0^{x}d\theta \frac{1}{2}{\left(2\sin \theta \right)}^2=2\sqrt{\frac{7}{5}}{\int }_0^x{\sin }^2\theta d\theta \\ =\sqrt{\frac{7}{5}}{\int }_0^x(1-\cos (2\theta \left)\right)d\theta \\ ={\overline{)\sqrt{\frac{7}{5}}\left[\theta -\frac{1}{2}\sin \left(2\theta \right)\right]}}_0^x \\ =\sqrt{\frac{7}{5}}\mathrm{\pi } \end{gathered}$$

The surface area cut from the cone $2x^2+2y^2=5z^2,z>0$by the cylinder $x^2+y^2=2y$is $A=\mathrm{\pi }\sqrt{\frac{7}{5}}.$