# How do you compute the volume of the solid formed by revolving the given the region bounded by #y=sqrtx, y=2, x=0# revolved about (a) the y-axis; (b) x=4?

I have chosen to do both parts by both shells and washers(discs). Therefore I have provided separate answers for questions (a) and (b). Here is part (b)

Part (b) by shells

For shells we'll take a slice parallel to the axis of rotation. SO that's a vertical slice at some value of

The radius of revolution is the distance between the slice and the axis of rotation. So,

As in part (a), the height is

The representative shell has volume

The volume of the resulting solid is

(Details are left to the reader.)

Part (b) by washers

Take the slice perpendicular to the axis of rotation. So we are taking a horizontal slice at a value of

The revolution results not in a disc, but in a washer.

The outer (greater) radius is

The representative washer has volume

(Details are left to the reader.)

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I have chosen to do both parts by both shells and washers(discs). Therefore I have provided separate answers for questions (a) and (b). Here is part (a)

Here is a graph of the region:

Part (a)

By Cylindrical Shells

We'll take a thin slice parallel to the axis of rotation -- in this case the

The location of the slice is

The volume of the representative shell is

In this case

So the representative shell has volume

The value of

(Details are left to the reader.)

Part (a) By discs (washers)

For discs or washers we take our slice perpendicular to the axiz of rotation.

In this case the thickness of the slice will be

The volume of the representative disc is

In this case

The volume of the representative disc is

The volume of the solid is

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To compute the volume of the solid formed by revolving the given region bounded by (y = \sqrt{x}), (y = 2), and (x = 0) about:

(a) the y-axis: Use the disk method. The volume (V) can be computed using the formula:

[ V = \pi \int_{a}^{b} [f(y)]^2 , dy ]

In this case, (a = 0) and (b = 4). The function (f(y)) represents the distance from the axis of revolution (y-axis) to the outer edge of the region. Since the region is bounded by (y = \sqrt{x}), (f(y) = \sqrt{x}). We need to solve for (x) in terms of (y): (x = y^2). Now, the integral becomes:

[ V = \pi \int_{0}^{2} (y^2)^2 , dy ]

[ V = \pi \int_{0}^{2} y^4 , dy ]

[ V = \pi \left[ \frac{1}{5}y^5 \right]_{0}^{2} ]

[ V = \frac{32}{5} \pi ]

(b) the line (x = 4): Use the shell method. The volume (V) can be computed using the formula:

[ V = 2\pi \int_{a}^{b} x \cdot f(x) , dx ]

In this case, (a = 0) and (b = 2). The function (f(x)) represents the distance from the axis of revolution (line (x = 4)) to the outer edge of the region. Since the region is bounded by (y = \sqrt{x}), (f(x) = 4 - \sqrt{x}). The integral becomes:

[ V = 2\pi \int_{0}^{2} x \cdot (4 - \sqrt{x}) , dx ]

[ V = 2\pi \left[ \frac{4}{2}x^2 - \frac{2}{3}x^{3/2} \right]_{0}^{2} ]

[ V = 2\pi \left( 4 - \frac{8}{3} \right) ]

[ V = \frac{16}{3} \pi ]

So, the volume of the solid formed by revolving the given region about (a) the y-axis is (\frac{32}{5} \pi) cubic units, and about (b) the line (x = 4) is (\frac{16}{3} \pi) cubic units.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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