# Tag Info

### 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 ...
• 80.9k
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### 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 ...
• 117k

### 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 ...
• 51.9k

### 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 ...
• 146k
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### 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)^...
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### 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 ...
• 618k
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### 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 ...
• 7,263
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### 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 $x$, ...
• 2,790

### 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 ...
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### 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 ...
• 139k
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### 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 ...
• 508k
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### 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 ...
• 7,605
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### 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\...
• 82.8k