How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y^{2} = 4 x$ $y^2=4x$ , x=y revolved about the y-axis?

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y^{2} = 4 x$ $y^2=4x$ , x=y revolved about the y-axis?

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Answer 1 · 2016-10-13T19:21:51

See below.

Explanation:

Draw a picture (sketch) of the region.
Note that the graphs intersect where $y = x$ $y=x$ and $y^{2} = 4 x$ $y^2=4x$ , so that is where $x^{2} = 4 x$ $x^2=4x$ .
And that happens at $(0, 0)$ $(0,0)$ and at $(4, 4)$ $(4,4)$ .

enter image source here

We are asked to use shells, so we take thin representative rectangles parallel to the axis of rotation. In this case that is the $y$ $y$ -axis.
The slice has solid black boundaries and the radius to the axis of rotation is shown as a dashed black line.

enter image source here

Since the thin part is $d x$ $dx$ , we need both equations as functions of $x$ $x$ .

$y^{2} = 4 x$ $y^2=4x$ in the first quadrant (which is where the region is) gets us $y = 2 \sqrt{x}$ $y = 2sqrtx$ for the upper function.

The lower function is $y = x$ $y=x$

The volume of a representative shell is $2 π r h \cdot thickness$ $2pirh*"thickness"$

In this case, we have

radius $r = x$ $r = x$ (the dashed black line),

height $h = u p p e r - l o w e r = 2 \sqrt{x} - x$ $h = upper - lower = 2sqrtx-x$ and

$thickness = d x$ $"thickness" = dx$ .

$x$ $x$ varies from $0$ $0$ to $4$ $4$ , so the volume of the solid is:

$\int_{0}^{4} 2 π x (2 \sqrt{x} - x) d x = 2 π \int_{0}^{4} (2 x^{\frac{3}{2}} - x^{2}) d x$ $int_0^4 2pi x(2sqrtx-x)dx =2 piint_0^4 (2x^(3/2)-x^2) dx$

(details left to the student)

$= \frac{128 π}{15}$ $= (128pi)/15$

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y^{2} = 4 x$ $y^2=4x$ , x=y revolved about the y-axis?

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Explanation:

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y^2=4xy2=4xy^2=4x, x=y revolved about the y-axis?

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Explanation:

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y^{2} = 4 x$ $y^2=4x$ , x=y revolved about the y-axis?