What is the volume of the solid formed using the shell method?

[karush]

https://www.physicsforums.com/attachments/1012

Let $$f(x)=\sqrt{x}$$, Line $$L$$ is the normal to the graph of f and point $$(4,2)$$

(a) show that the equation of L is $$y=-4x+18$$

$$f'=\frac{1}{2\sqrt{x}}$$ so $$f'(4)=\frac{1}{4}$$ so normal would be $$-4$$
then L is $$y-2=-4(x-4)$$ or $$y= -4x+18$$



(b) Point $$A$$ is the x-intercept of $$L$$. find the x-coordinate of $$A$$

from $$Y=-4x+18$$ set $$0=-4x+18$$ then $$18=4x$$ and $$x=\frac{9}{2}$$ so $$A=(\frac{9}{2},0)$$

there is still (c) and (d) but want to make sure this is correct first.

 

[I like Serena]

Re: volumn of the solid formed

Yeah, yeah. This is correct.

Edit: that is, (a) is correct.
You've added (b) to the OP, which is still correct.

 

[karush]

Re: volumn of the solid formed

(c) find the expression for the area of R

$$\int_{0}^{4}\sqrt{x}\,dx + \int_{4}^{\frac{9}{2}}(-4x+18)\,dx$$

$$= \frac{16}{3} + \frac{1}{2} = \frac{35}{6}$$

(d) The region R is rotated $$360^0$$ about the $$x$$-axis. Find the volume of the solid formed
give answer in terms of $$\pi$$

I assume this means adding $$2$$ integrals

$$\pi\bigg[\int_{0}^{2\pi}(\sqrt{x})^2\,dx +\int_{0}^{2\pi}(-4x+18)^2\,dx\bigg]\approx 536\pi$$ units$$^3$$

 

[MarkFL]

Re: volumn of the solid formed

c) Correct.

d) Check your limits of integration. They should be in terms of $x$, not $\theta$.

 

[I like Serena]

Re: volumn of the solid formed

(c) find the expression for the area of R

$$\int_{0}^{4}\sqrt{x}\,dx + \int_{4}^{\frac{9}{2}}(-4x+18)\,dx$$

$$= \frac{16}{3} + \frac{1}{2} = \frac{35}{6}$$

(d) The region R is rotated $$360^0$$ about the $$x$$-axis. Find the volume of the solid formed
give answer in terms of $$\pi$$

I assume this means adding $$2$$ integrals

$$\pi\bigg[\int_{0}^{2\pi}(\sqrt{x})^2\,dx +\int_{0}^{2\pi}(-4x+18)^2\,dx\bigg]\approx 536\pi$$ units$$^3$$

Looking good.
I guess you just wanted to verify you're using the right method?

Edit: as Mark said, you need to rectify your integral boundaries.

 

[karush]

Re: volumn of the solid formed

Looking good.
I guess you just wanted to verify you're using the right method?

Edit: as Mark said, you need to rectify your integral boundaries.

$$\pi\bigg[\int_{0}^{4}(\sqrt{x})^2\,dx +\int_{4}^{\frac{9}{2}}(-4x+18)^2\,dx\bigg]\approx 8.67\pi$$ units$$^3$$

how does this assume that we are going $$360^0$$ around the $$x$$ axis if $$2\pi$$ isn't anywhere?

 

[I like Serena]

Re: volumn of the solid formed

$$\pi\bigg[\int_{0}^{4}(\sqrt{x})^2\,dx +\int_{4}^{\frac{9}{2}}(-4x+18)^2\,dx\bigg]\approx 8.67\pi$$ units$$^3$$

how does this assume that we are going $$360^0$$ around the $$x$$ axis if $$2\pi$$ isn't anywhere?

You're integrating circle disks.
Each disk has surface $\pi r^2$, which effectively includes your $2\pi$.
And each disk has a thickness of $dx$, which is what you are integrating.

 

[MarkFL]

Re: volumn of the solid formed

$$\pi\bigg[\int_{0}^{4}(\sqrt{x})^2\,dx +\int_{4}^{\frac{9}{2}}(-4x+18)^2\,dx\bigg]\approx 8.67\pi$$ units$$^3$$

how does this assume that we are going $$360^0$$ around the $$x$$ axis if $$2\pi$$ isn't anywhere?

Your integrands are of the form $\pi r^2$ which combined with the differential, gives the volumes of a stack of disk shaped slices (complete circular disks, thus the angle of rotation of $360^{\circ}$ is implied) which are then summed up by integration.

 

[karush]

Re: volumn of the solid formed

Your integrands are of the form $\pi r^2$ which combined with the differential, gives the volumes of a stack of disk shaped slices (complete circular disks, thus the angle of rotation of $360^{\circ}$ is implied) which are then summed up by integration.

ok got it...
so my volume is correct then...

 

[MarkFL]

Re: volumn of the solid formed

ok got it...
so my volume is correct then...

I would refrain from using a decimal approximation for the factor of $\pi$. I would write:

$$V=\frac{26\pi}{3}$$

As a follow-up...can you compute the volume using the shell method? This will actually allow the volume to be computed using only 1 integral.(Nerd)