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Find the area of the plane x-2y+5z=13 cut out by the cylinder x2+y2=9.

Short Answer

Expert verified

The area of the plane x-2y+5z=13cut out by the cylinder x2+y2=9 isA=9Ï€305.

Step by step solution

01

Given Condition

The equation of the plane is x-2y+5z=13 and the equation of the cylinder is x2+y2=9.

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

04

Calculate the angle

The secant of the angle γis given by,

secγ=∇ϕ2∂ϕ/∂z=x^-2y^+5z^25=1+4+55=305

05

Calculate the area

The area of the plane cut by the cylinder can be found by,

A=∫x∫ysecγdxdy=305(9π)=9π305

Hence, the required area is A=9Ï€305.

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