Find the area of the part of the cylinder${\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{=}\mathbf{4}$in the first octant, cut out by the planes$\mathbf{x}\mathbf{=}\mathbf{0}$and$\mathbf{y}\mathbf{=}\mathbf{x}$

The equation of the cylinder is${\mathrm{y}}^2+{\mathrm{z}}^2=4$

The planes are $x=0$and$y=x$ .

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 shaded region is the region of integration.

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The equation of the cylinder is $\varphi =y^2+z^2$

The secant of the angle is found as follows.

role="math" $$\begin{gathered} sec\gamma =\frac{\sqrt{{\left|\nabla \varphi \right|}^2}}{\left|\partial \varphi /\partial z\right|} \\ =\frac{\sqrt{{\left|2y\hat{y}+2z\hat{z}\right|}^2}}{2z} \\ =\frac{\sqrt{y^2+z^2}}{z} \\ =\frac{\sqrt{y^2+4-y^2}}{\sqrt{4-y^2}} \\ =\frac{2}{\sqrt{4-y^2}} \end{gathered}$$

The required area is calculated as below.

$$\begin{gathered} A={\int }_x{\int }_{y}dxdysec\gamma \\ ={\int }_0^{2}dx{\int }_x^{2}dy\frac{2}{\sqrt{4-y^2}} \\ puty=2\sin u \\ \Rightarrow dy=2\cos udu \\ A=2{\int }_0^{2}dx{\int }_{arc\sin (x/2)}^{x/2}\frac{2\cos udu}{\sqrt{4-4{\sin }^{2}u}} \\ =2{\int }_0^{2}dx{\int }_{arc\sin (x/2)}^{x/2}du \\ =2{\int }_0^{2}dx\left(\frac{\mathrm{\pi }}{2}-arc\sin \frac{x}{2}\right) \\ =2\left[\mathrm{\pi }-{\int }_0^2\arcsin \frac{\mathrm{x}}{2}\mathrm{dx}\right] \\ ={\overline{)2\mathrm{\pi }-2\left(\sqrt{4-{\mathrm{x}}^2}+\mathrm{x}\arcsin \frac{\mathrm{x}}{2}\right)}}_0^2 \\ =2\mathrm{\pi }-2(0+\mathrm{\pi }-2) \\ =4 \end{gathered}$$

Hence, the area of the part of the cylinder $y^2+z^2=4$in the first octant, cut out by the planes $x=0$and $y=x$is$A=4.$