Question #176683

Use the method disks/rings to determine the volume of the solid obtained by rotating the region bounded by the given curves about the given axis of y=2x+1, x=4 and y= 3 about the line x=-4.


1
Expert's answer
2021-04-14T02:43:55-0400

The given lines are represented as-




The region formed by rotating the above regions is-



So, the inner and outer radii are as follows:


Inner Radius r=4+12(y1)=y2+72r=4+\dfrac{1}{2}(y-1)=\dfrac{y}{2}+\dfrac{7}{2}


Outer radius R=4+4=8R=4+4=8


Area of the ring is

A(x)=π[R2r2]A(x)=\pi [R^2-r^2]


   =π[(8)2(y2+72)2]=\pi[(8)^2-(\dfrac{y}{2}+\dfrac{7}{2})^2]


   =π(207472yy24)=\pi(\dfrac{207}{4}-\dfrac{7}{2}y-\dfrac{y^2}{4})



Then the volume of the solid obtained is


 V=39A(x)dy=π(207472yy24)=39π(207472yy24)dyV=\int_3^9A(x)dy=\pi(\dfrac{207}{4}-\dfrac{7}{2}y-\dfrac{y^2}{4})=\int_3^9 \pi(\dfrac{207}{4}-\dfrac{7}{2}y-\dfrac{y^2}{4})dy


  =π(207y47y24y312)39=126π=\pi(\dfrac{207y}{4}-\dfrac{7y^2}{4}-\dfrac{y^3}{12})|_3^9=126\pi


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