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?

Question

Isaiah Allen · Accepted Answer

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 #x# and the thickness is #dx#.

The radius of revolution is the distance between the slice and the axis of rotation. So, #r = 4-x#

As in part (a), the height is #h = y_"top"-y_"bottom" = 2-sqrtx#

The representative shell has volume

#2pirh*"thickness" = 2pi(4-x)(2-sqrtx)dx#

#x# varies from #0# to #4#.

The volume of the resulting solid is

#int_0^4 2pi(4-x)(2-sqrtx)dx = (224pi)/15#
(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 #y#. The curve involves has equation #x = y^2# (in the first quadrant).

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

The outer (greater) radius is #R = 4-x_"left" = 4-0 = 4# and the inner (lesser) radius is #r = 4-x_"right" = 4-y^2#

The representative washer has volume

#pi(R^2-r^2)*"thickness" = pi(4^2-(4-y^2)^2)dy#

#y# varies from #0# to #2#, so the solid has volume

#int_0^2 pi(4^2-(4-y^2)^2)dy =(224pi)/15#
(Details are left to the reader.)

Ezekiel Alexander · Answer

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 #y#-axis. The thickness of the slice will be #dx#. So we will be integrating with respect to #x#.

The location of the slice is #x#, so the radius of the cylinder (the dotted black line) is #x-0 = x#

The volume of the representative shell is #2pirh*"thickness"#

In this case #r = x# and

#h = y_"top"-y_"bottom"# (the greater value of #y# minus the lesser) which is #h = (2-sqrtx)#

#"thickness" = dx#

So the representative shell has volume #2pix(2-sqrtx)dx#

The value of #x# go from #0# to #4#, so the volume of the solid of revolution is

#int_0^4 2pix(2-sqrtx)dx = 2piint_0^4 (2x-x^(3/s))dx = (32pi)/5#
(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 #dy# so we'll be integrating with respect to #y#. The slice is taken at a value of #y#.

The volume of the representative disc is #pir^2*"thickness"#

In this case #r = x# at our chosen value of #y#. But we need #r# in terms of #y#. The curve #y = sqrtx# is the upper part of #x=y^2#. (We have only positive values of #y#.) So,

#r = y^2#.

The volume of the representative disc is #pi(y^2)^2dy#

#y# varies from #0# to #2#

The volume of the solid is

#int_0^2piy^4dy = (32pi)/5#

Elijah Abram · Answer

undefined 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.