Find the volume of the solid that results when the region enclosed by `x=y^2` and x=y is revolved about the line y=-3.

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valentin68 | College Teacher | (Level 3) Associate Educator

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In the figure attached above there is a small mistake: the area of the disk shown is `pi*(x+3)^2` not `pi*(x-3)^2`

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valentin68 | College Teacher | (Level 3) Associate Educator

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For simplicity we will exchange x and y coordinates and find the area of the figure enclosed between `y=x^2` and `y=x` when revolving around `x=-3` .

The equation `x^2 =x` has two real roots, `x_1=0` and `x_2=1` so that the intersection points of the two given curves (or functions) are 0 and 1.

If we rotate the area between `f(x) =x^2` and `g(x) =x` about the axis `x=0` the volume of rotation can be found by applying the disc method

`V_1 =pi*int_0^1[g(x)^2-f(x)^2]dx =pi*int_0^1[x^2 -x^4]dx =pi*(X^3/3 -X^5/5) (0->1) =pi*(1/3 -1/5) =pi*0.133=0.4189`

When rotating the given area around the axis `x=-3` all you do is to add at the radius of disks that we sum the constant `x=3` . Thus the radius of disks will be `f(x)+3` and `g(x)+3` .

Therefore the volume of rotation will become

`V_2 =pi*int_0^1[(g(x)+3)^2 -(f(x)+3)^2]dx =pi*int_0^1[(x+3)^2 -(x^2+3)^2 ]dx=`

`=pi*int_0^1[x^2+6x+9 -x^4-6x^2 -9]dx =pi*int_0^1[-x^4 -5x^2+6x]dx =`

`=pi*(-x^5/5 -5/3*x^3 +3x^2)ยด(0->1) =pi*(-1/5 -5/3 +3) =pi*1.133 =3.56`

Thus the volume of the figure obtained by rotating the area between `x=y^2` and `x=y` around `y=-3` is 3.56

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