Saturday 31 October 2015

`y = 1 + sec(x), y = 3` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

`y=1+sec(x),y=3`


Refer the image. From the graph, the curves intersects at x=-pi/3 and x=pi/3.


Using washer method,


A cross section is a washer of cross sectional area A(x) with,


Inner radius=(1+sec(x))-1=sec(x)


Outer radius=3-1=2


`A(x)=pi(2^2-(sec(x))^2)`


`A(x)=pi(4-sec^2(x))`


Volume of the solid obtained by rotating the region bounded by the given curves about y=1 (V) is,


`V=int_(-pi/3)^(pi/3)A(x)dx`


`V=int_(-pi/3)^(pi/3)pi(4-sec^2(x))dx`


`V=2piint_0^(pi/3)(4-sec^2(x))dx`


`V=2pi[4x-tan(x)]_0^(pi/3)`


`V=2pi((4*pi/3-tan(pi/3))-(4*0-tan(0)))`


`V=2pi(4*pi/3-sqrt(3)-0)`


`V=2pi((4pi)/3-sqrt(3))`


`y=1+sec(x),y=3`


Refer the image. From the graph, the curves intersects at x=-pi/3 and x=pi/3.


Using washer method,


A cross section is a washer of cross sectional area A(x) with,


Inner radius=(1+sec(x))-1=sec(x)


Outer radius=3-1=2


`A(x)=pi(2^2-(sec(x))^2)`


`A(x)=pi(4-sec^2(x))`


Volume of the solid obtained by rotating the region bounded by the given curves about y=1 (V) is,


`V=int_(-pi/3)^(pi/3)A(x)dx`


`V=int_(-pi/3)^(pi/3)pi(4-sec^2(x))dx`


`V=2piint_0^(pi/3)(4-sec^2(x))dx`


`V=2pi[4x-tan(x)]_0^(pi/3)`


`V=2pi((4*pi/3-tan(pi/3))-(4*0-tan(0)))`


`V=2pi(4*pi/3-sqrt(3)-0)`


`V=2pi((4pi)/3-sqrt(3))`


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