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Find the area of the part of the cylindery2+z2=4in the first octant, cut out by the planesx=0andy=x

Short Answer

Expert verified

The area of the part of the cylinder y2+z2=4in the first octant, cut out by the planes x=0and y=xisA=4 .

Step by step solution

01

Given Condition

The equation of the cylinder isy2+z2=4

The planes are x=0andy=x .

02

Concept of surface area

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.

03

Draw the diagram

The shaded region is the region of integration.

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04

Calculate the angle

The equation of the cylinder is ϕ=y2+z2

The secant of the angle is found as follows.

role="math" secγ=ϕ2ϕ/z=2yy^+2zz^22z=y2+z2z=y2+4-y24-y2=24-y2

05

Calculate the area.

The required area is calculated as below.

A=xydxdysecγ=02dxx2dy24-y2puty=2sinudy=2cosuduA=202dxarcsin(x/2)x/22cosudu4-4sin2u=202dxarcsin(x/2)x/2du=202dxπ2-arcsinx2=2π-02arcsinx2dx=2π-24-x2+xarcsinx202=2π-2(0+π-2)=4

Hence, the area of the part of the cylinder y2+z2=4in the first octant, cut out by the planes x=0and y=xisA=4.

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