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[SOLVED] Volume of the solid formed

karush

Well-known member
Jan 31, 2012
2,779
View attachment 1012

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

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

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



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

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

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,900
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.
 
Last edited:

karush

Well-known member
Jan 31, 2012
2,779
Re: volumn of the solid formed

(c) find the expression for the area of R

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

\(\displaystyle = \frac{16}{3} + \frac{1}{2} = \frac{35}{6}\)

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

I assume this means adding \(\displaystyle 2\) integrals

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: volumn of the solid formed

c) Correct.

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,900
Re: volumn of the solid formed

(c) find the expression for the area of R

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

\(\displaystyle = \frac{16}{3} + \frac{1}{2} = \frac{35}{6}\)

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

I assume this means adding \(\displaystyle 2\) integrals

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

Well-known member
Jan 31, 2012
2,779
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.
\(\displaystyle \pi\bigg[\int_{0}^{4}(\sqrt{x})^2\,dx +\int_{4}^{\frac{9}{2}}(-4x+18)^2\,dx\bigg]\approx 8.67\pi\) units\(\displaystyle ^3\)

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,900
Re: volumn of the solid formed

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

how does this assume that we are going \(\displaystyle 360^0\) around the \(\displaystyle x\) axis if \(\displaystyle 2\pi\) 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

Administrator
Staff member
Feb 24, 2012
13,775
Re: volumn of the solid formed

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

how does this assume that we are going \(\displaystyle 360^0\) around the \(\displaystyle x\) axis if \(\displaystyle 2\pi\) 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

Well-known member
Jan 31, 2012
2,779
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

\(\displaystyle V=\frac{26\pi}{3}\)

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)