Skip to main content
9 votes

Why do we care so much about solids/surfaces of revolution?

One answer is certainly that these are objects that we can analyze using single-variable calculus techniques, and so textbooks and curriculums include them because they can be done. But I want to add ...
Greg Martin's user avatar
  • 80.9k
8 votes
Accepted

Convert Surface of Revolution to Parametric Equations

Any surface of revolution can be easily parametrized. If you start with the parametric curve $(x(u),y(u))$, $u\in I$ (some interval), and rotate it about the $x$-axis, the surface you obtain will be ...
Ted Shifrin's user avatar
8 votes

Disc vs Shell Method, getting different answers AP calc

Your working for the disk method is correct. In shell method, the integral should be, $\displaystyle 2 \pi \int_{1/e}^{1} \color {blue} {(1-y)} (e-\frac{1}{y}) \ dy$ The mistake that you made was in ...
Math Lover's user avatar
  • 51.9k
6 votes

Calculate the volume of a torus

By Pappus's centroid theorem, the volume of the torus is given by $2\pi R\cdot \pi r^2$. The same result can be obtained by using integration and cylindrical coordinates: the torus is generated by the ...
Robert Z's user avatar
  • 146k
5 votes
Accepted

Solid of Revolution About y=2

Hint: What you are looking for Your answer \begin{align*} \text{Volume }&=\text{Volume of the solid without the "hole"}-\text{Volume of the cylinder }\\ &=\pi\displaystyle\int_{-1}^1(2-x^4)^...
Ángel Mario Gallegos's user avatar
5 votes
Accepted

Shell Method Versus Disc Method

The solid whose volume you calculated using the disk method is not the solid whose volume you calculated using the shell method. If you used shells on the first solid, your shells would run from the ...
Brian M. Scott's user avatar
5 votes
Accepted

Solid of revolution of $\frac{1}{x^a}$

There's a mistake on the step $$ \int_0^1\frac{1}{x^{2a}} dx = \frac{1^{-2a+1}}{-2a+1} $$ which is not always true. To see why we'll split the problem in cases. If $\color{blue}{a<\frac{1}{2}}$ we ...
Robert Lee's user avatar
  • 7,263
5 votes
Accepted

How do i solve this integration question using the washer and shell method?

Washer Method The larger radius comes from the right side of the parabola $y = (x - 1)^2$, while the smaller radius comes from the left side of that parabola. Rewriting that parabola in terms of $x$, ...
Amaan M's user avatar
  • 2,790
4 votes

Calculate surface area of revolution with arc length formula

After carefully reviewing your problem, I've identified a small misunderstanding that might have led to the confusion. It seems like you've taken $y$ out of the integral formula, which is a common ...
Arthur Kangdani's user avatar
4 votes
Accepted

Set up a definite integral for the volume obtained by rotating the region between the curve $y^{2}=x$ and $y^{2}=2(x-1)$ about $y=3$

Using the method of washers, the washers would be perpendicular to the axis of rotation, thus the integration would be performed along the $x$-axis. However, because a line perpendicular to the axis ...
heropup's user avatar
  • 139k
4 votes
Accepted

Difficulty setting up a solid of revolution integral

The cross-section is obtained by removing a disk of radius $y^2/8$ from a disk of radius $2$. That gives volume $$\int_{-4}^4 \pi \cdot(2^2)\,dy-\int_{-4}^4 \pi \cdot\frac{y^4}{64}\,dy.$$ The first ...
André Nicolas's user avatar
4 votes
Accepted

Apostol Vol.1 Chap. 2 execises 2.13 n°15

The answer that you calculated is incorrect in that you have assumed that the solid is a cone. In fact, it is not. A cone does not have similar triangles in its cross sections; they are hyperbolas. We ...
Cye Waldman's user avatar
  • 7,605
4 votes
Accepted

Volume generated by revolving the region in the first quadrant bounded by the parabolas $y^2=x$, $y^2=8x$, $x^2=y$, $x^2=8y$ about the $x$-axis.

Your first term is correct, but the second and third are incorrect. The correct volume expression should read $$V = \pi \left(\int_{x=1}^2 (\color{magenta}{x^2})^2 - (\color{orange}{\sqrt{x}})^2 \, ...
heropup's user avatar
  • 139k
4 votes
Accepted

Find the volume of the solid obtained by rotating the region bounded by the curves $y=x^2$, $x=5$, and $y=0$ about the $x$-axis

Your bounds of integration are going to be from $0$ to $5$. Your radius function is going to be $r(x)=x^2$. Now, just plug it into the formula for the disk method which is $V=\pi\int_{a}^{b}\left[r(x)\...
Michael Rybkin's user avatar
4 votes

Volume of revolution found by rotating the region bounded by $x=y^2$ and $x=1-y^2$ about the line $x=3$

There are two basic approaches to this. One is to integrate with respect to $x$, the other with respect to $y$. In both cases, for each value of our chosen variable we find a relevant area, and then ...
Arthur's user avatar
  • 201k
4 votes

What is the correct formula for the washer method?

The first equation is correct and the second wrong. Just study this figure where $f(x) - g(x)$ is the same for the red and blue. Your second equation incorrectly says they have the same area.
David G. Stork's user avatar
4 votes
Accepted

What is the correct formula for the washer method?

The first formula is correct, subject to confusion that most books (that I use to teach, and that are not my choice) do not clearly explain what are these $f(x)$ and $g(x)$. (Either that, or students ...
Mirko's user avatar
  • 13.5k
4 votes
Accepted

Pappus centroid theorem and Hypercones.

tl; dr: The formulas work out for a cone of height $h$ and base radius $R$ in four-space. The volume is indeed \begin{align*} \tfrac{1}{3}\pi R^{3}h &= (\tfrac{1}{2}Rh)(\tfrac{2}{3}\pi R^{2}) \\ ...
Andrew D. Hwang's user avatar
4 votes

Volumes of Rotation: Where's my mistake??

You seem to have made a typo in writing the first function: $y = (-x^2 + 6x - 5)^2$, and I shall proceed by assuming that you meant $y = (-x^2 + 6x - 5)$. For a rotation around the line $y=h$, the ...
allegro.sostenuto's user avatar
4 votes

Volume obtained by rotating the area enclosed by $y=x^2$ and $y=\sqrt x$ about $y=1$ vs about $x$-axis

I haven't checked your calculations (and I suppose the volumes might actually be the same, if you simply made an error). But you say: By looking at the area, volumes look the same. and I think this ...
3 votes
Accepted

Volume of Solid of Revolution (Disk/Washer)

The intersection between the curves $x=y-y^2$ and $x=0$ is obtained by considering the equation $$0=y-y^2.$$ Solving the equation, we get $y=0$ or $y=1$. The points of intersection are $(0,0)$ and $(0,...
Juniven Acapulco's user avatar
3 votes
Accepted

Pappus's Centroid Theorem

$\newcommand{\dd}{\partial}$If $\bigl(x(t), z(t)\bigr)$, $a \leq t \leq b$, parametrizes a smooth plane curve $C$ in the half-plane $x > 0$, the surface $S$ swept out by revolving $C$ about the $z$-...
Andrew D. Hwang's user avatar
3 votes

Volume of Revolution: Volume of resulting solid

if you consider a thin disc of thickness $dy$ & radius $r=x(y)+3$ about the line $x=-3$ & subtract the volume of corresponding thin concentric disc $\pi (3)^2dy$ then the required volume of ...
jeanne clement's user avatar
3 votes

Centroid of a solid of revolution

The centroid of any volume is defined by $$ \vec{c} = \frac{ \int \vec{r} \,{\rm d}V}{ \int {\rm d}V} $$ For a volume of revolution about the x-axis ${\rm d}V = r\, {\rm d}\theta\, {\rm d}r \, {\rm ...
John Alexiou's user avatar
  • 14.1k
3 votes

Volume of Revolution Help

If you plot out the graph, you will see that for the bounded region, the linear line is above the exponential function. You have used a smaller quantity to subtract a bigger quantity, hence, you get a ...
Siong Thye Goh's user avatar
3 votes

Calculate the volume of a torus

Using cylindrical coordinates (As $r$ is usually used in this coordinates I will use $S$ instead): by symmetry, $(x-R)^2 + z^2 = S^2$ gives $$ (r-R)^2 + z^2 = S^2 \implies r = R\pm\sqrt{S^2-z^2},\...
Martín-Blas Pérez Pinilla's user avatar
3 votes
Accepted

Calculating the surface of revolution of a cardioid.

First a tip: it is often easier to substitute back for $r$ in terms of $t$ after differentiating. We have $$\displaylines{ x=r\cos t\ ,\quad y=r\sin t\cr \frac{dx}{dt}=\frac{dr}{dt}\cos t-r\sin t\...
David's user avatar
  • 82.8k
3 votes
Accepted

If I have an oblique cylinder can I trim it in to a rectilinear cylinder?

If you mean by an oblique cylinder a surface obtained by shearing a cylinder, then the answer is in general NO. Its cross-section perpendicular to the 'axial direction' will be an ellipse, which means ...
Sangchul Lee's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible