91Ó°ÊÓ

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 $\mathrm{Ï•}=y^2+z^2$

The secant of the angle is found as follows.

role="math" $$\begin{gathered} sec\mathrm{γ}=\frac{\sqrt{{\left|∇\mathrm{ϕ}\right|}^2}}{\left|∂\mathrm{ϕ}/∂z\right|} \\ =\frac{\sqrt{{\left|2y\hat{y}+2z\hat{z}\right|}^2}}{2z} \\ =\frac{\sqrt{y^2+z^2}}{z} \\ =\frac{\sqrt{y^2+4-y^2}}{\sqrt{4-y^2}} \\ =\frac{2}{\sqrt{4-y^2}} \end{gathered}$$

The required area is calculated as below.

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

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$A=4.$