Find the volume of the solid generated by revolving the region about the given line.The region in the first quadrant ounded above by the line y=2, below by the curve y=sqrt(2x), and on the left by...

Find the volume of the solid generated by revolving the region about the given line.

The region in the first quadrant ounded above by the line y=2, below by the curve y=sqrt(2x), and on the left by the y-axis, about the line x=-1

The answe is 64pi/15, but I cannot get the correct answer. Can anyone show me the process??

Asked on by dylachan

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You must notice that you are about to find the volume of the walls of the solid, excluding the inner part of this solid.

You need to remember the formula of the volume of solid of revolution: `V = (2pi*r)*h`

`r = x+1`  denotes the radius of the solid of revolution measured from y axis to the curve `y = 2 - sqrt (2x)`

`h = 2 - sqrt (2x)`  denotes the height of the solid of revolution

You need to find the points of intesection of the curves bounding the solid of revolution:

Equating the given relations yields:

`2 = sqrt 2x`

Raising to square to remove the square root yields:

`4 = 2x =gt x = 2`

Since the region is found in the first quadrant, the bounding limits are (0,2).

Hence, the volume of rotation is:

`V = 2pi* int_0^2(2 - sqrt (2x))*(x+1)dx`

`V = 2pi*(int_0^2 2xdx + int_0^2 2dx - int_0^2 sqrt2*x^(1/2+1)dx - int_0^2 sqrt2*x^(1/2))`

`V = 2pi (x^2 + 2x - (2sqrt2/5)*x^(5/2) - (2sqrt2/3)*x^(3/2))_0^2`=>`V = 2pi (4 + 4 - 16/5 -8/3)`

`V = 2pi(15*8 - 48 - 40)/15 =gt V= 2pi*32/15 =gt V = 64pi/15`

The volume of solid of revolution is of `64pi/15` .

embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

We use the shell method where `V=2pi int_a^bp(x)h(x)dx` . p(x) is the distance of the rectangle from the vertical axis of rotation, and h(x) is the height of the rectangle.

The two curves intersect at (2,2), thus the limits of integration are 0 and 2.

p(x)=x+1 (The distance from x=-1 as x goes from 0 to 2)

`h(x)=2-sqrt(2x)` since the line y=2 is always above `y=sqrt(2x)` on (0,2). We take the area under y=2 and subtract the area under the square root curve.

Then `V=2pi int_0^2(2-sqrt(2x))(1+x)dx`

`=` `2pi int_0^2(2+2x-(2x)^(1/2)-x(2x)^(1/2))dx`

`=2pi[2x+x^2-1/3(2x)^(3/2)-sqrt(2)(2/5)x^(5/2)|_0^2`

`=2pi(4+4-8/3-(16)/5)`

`=(64pi)/(15)`

 

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