S4.6r.11 Solid of revolution about the y axis

[karush]

Given
$$x_1^2 -y^2=a^2, \ \ x_2=a+h$$
Or
$$x_1=\sqrt{a^2+y^2}$$

Find
Volume about the $y$-axis

So...

$$\pi\int_{a}^{h} \left(x_2^2-x_1^2\right)\,dy$$

Actually I am clueless?!

 

[MarkFL]

Is that how the problem is given in your textbook?

 

[MarkFL]

Well, that's clear as floodwaters, my friend! (Nod)

Can you humor me for a little bit here and give the problem exactly as stated? (Mmm)

 

[karush]

View attachment 5538
It's #11

 

[MarkFL]

Okay, now we're in bidness...and your inclination to use the washer method is a good one...but you know, when I was a student, I like to work problems like these more than one way (we could use the shell method as well) both for the prctice and as a means of checking my result.

Let's begin with the washer method. The value of an element (a washer) is given by:

\(\displaystyle dV=\pi(R^2-r^2)\)

where $R$ is the outer radius and $r$ is the inner radius. Can you identify these two radii?

 

[karush]

I assume $R$ is the major radius or $x_2$
And $r$ is the minor radius or $x_1$

No I'm not sure?

This thing has $a$ and $h$ in it

 

[MarkFL]

I assume $R$ is the major radius or $x_2$
And $r$ is the minor radius or $x_1$

No I'm not sure?

This thing has $a$ and $h$ in it

First, it is often very helpful to draw a sketch:

View attachment 5539

Now, we see the outer radius is:

\(\displaystyle R=a+h\)

and the inner radius is:

\(\displaystyle r=x\)

And so we have:

\(\displaystyle dV=\pi\left((a+h)^2-x^2\right)\,dy\) (I forgot to include the thickness of the washer before)

Now, the expression $a+h$ is a constant, so we don't need to do anything with that. However, since we will be integrating along the $y$-axis, we need to express $x^2$ in terms of $y$...(Thinking)

 

[karush]

Sorry,I got lost in this,

The bk ans is $$\frac{4}{3}(2ah+h^2)^{3/2}$$

However $$x=\sqrt{y}$$

 

[MarkFL]

Since we are given:

\(\displaystyle x^2-y^2=a^2\)

we then know:

\(\displaystyle x^2=y^2+a^2\)

Hence:

\(\displaystyle dV=\pi\left((a+h)^2-\left(y^2+a^2\right)\right)\,dy\)

Now, you need to find the limits of integration, which will be the $y$-coordinates of the intersections of the line:

\(\displaystyle x=a+h\)

and the curve:

\(\displaystyle x^2-y^2=a^2\)

Can you proceed? :)

 

[karush]

$$dV=\pi\left((a+h)^2-\left(y^2+a^2\right)\right)\,dy$$dV
$$\pi\int_{-a}^{a}\left((a+h)^2-\left(y^2+a^2\right)\right) \,dy$$

Don't know what $h$ is?

 

[MarkFL]

I get different limits of integration. we have:

\(\displaystyle y^2=x^2-a^2\)

Now, substituting $c=a+h$, we obtain:

\(\displaystyle y=\pm\sqrt{h(2a+h)}\)

And then using the even function rule, we may write:

\(\displaystyle V=2\pi\int_0^{\sqrt{h(2a+h)}} \left((a+h)^2-\left(y^2+a^2\right)\right)\,dy\)

$h$ is just a constant, that is greater than the constant $a$. :)