# [SOLVED]Volume of the solid formed

#### karush

##### Well-known member
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.

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

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#### karush

##### Well-known member
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

Staff member
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
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
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
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

Staff member
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
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

I would refrain from using a decimal approximation for the factor of $\pi$. I would write:
$$\displaystyle V=\frac{26\pi}{3}$$