Volume of Region Bounded by x^2-y^2=16, y=0, x=8, about y-axis

[Dethrone]

Region bounded by $$x^2-y^2=16, y=0, x=8$$, about y-axis

Integrating with respect to y, my lower and upper bounds are $-4\sqrt{3}$ and $+4\sqrt{3}$, respectively.

$$V=\pi \int_{-4\sqrt{3}}^{4\sqrt{3}} 64-(16+y^2)\,dy$$
$$=2\pi \int_{0}^{4\sqrt{3}} 48-y^2\,dy$$
$$=2\pi (192\sqrt{3})$$
$$=256\pi$$

My textbook answer says the answer is $128\pi$, but I have a hunch that it's wrong because it probably also assumed that the curve was bounded by x=0, which it is not.

Not sure if this is a new feature, but I included the graph in spoilers:

Spoiler

[desmos="-10,10,-10,10"]x^2-y^2=16;x=8[/desmos]

 

[MarkFL]

Your answer is double that of your textbook because you did not account for the bounding by $y=0$. :D

 

[Dethrone]

If I'm integrating w.r.t y, how can I incorporate the bound y=0 in my calculations? Or will I just have to divide my answer by 2?

I guess I could maybe use a cylindrical shell method using x-bounds to eliminate the issue, but the question specifically asked for the washer method.

 

[MarkFL]

If I'm integrating w.r.t y, how can I incorporate the bound y=0 in my calculations? Or will I just have to divide my answer by 2?

I guess I could maybe use a cylindrical shell method using x-bounds to eliminate the issue, but the question specifically asked for the washer method.

You would simply use $y=0$ as your lower limit of integration. :D

 

[Dethrone]

Whoops! For some reason, I got $$x=0$$ and $$y=0$$ mixed up. $$y=0$$ is the horizontal line...and I was thinking it was the vertical y-axis the whole time...(Headbang)(Headbang)(Headbang)(Headbang)(Headbang)

 

[Dethrone]

Another related problem:

The region bounded by $y=x^3$, $x=0$, $y=8$, about $x=2$.

Is it referring to the region above the curve, or below the curve?
[desmos="-10,10,-10,10"]y=x^3;x=0;y=8;x=2[/desmos]

 

[MarkFL]

It is referring to the region above the cubic, below the horizontal line and to the right of the $y$-axis.

 

[Dethrone]

In which case, wouldn't rotation that about $x=2$ have the same volume as if you were to rotate that about the y-axis?

Constructing washers, we see that the area of a washer:
$$A=\pi [(r_o)^2-(r_i )^2]$$
$$A=\pi [(2)^2-(2-(f(y) )^2]$$
$$A=\pi [(2)^2-(2-y^{1/3} )^2]$$
$$A=\pi [4y^{1/3}-y^{2/3}]$$
$$V=\pi [4y^{1/3}-y^{2/3}] dy$$

Calculating the volume of all the washers:

$$V=\pi \int_{0}^{8}[4y^{1/3}-y^{2/3}] \,dy$$
$$=\frac{144\pi}{5}$$

I've answered my own question, no, they wouldn't have the same area. I'm having a bad day today, (Rain). I did it the right way the first time but made an arithmetic error, forcing me to believe that the area they're looking for was under $y=x^3$. Thanks again, Mark!