# Thread: Don't understand these problems?

1. ## Don't understand these problems?

[IMG]file:///C:/Users/c/AppData/Local/Temp/moz-screenshot.png[/IMG]

This problem I do not get at all...

2nd problem. I did all this

Center is (R,0)
r² = (x-R)² + y² ---eqn of a circle at (R,0)
y = ±√ r² - (x-R)²
f(x) = ±√ r² - (x-R)²

h = √ (r² - (x-R)²) - -√( r² - (x-R)²)
h = 2 √ (r² - (x-R)² )

Interval [a,b ] is [R-r,R+r]

V = 2π∫ x h dx [a,b]
V = 2π∫ x [2 √ (r² - (x-R)² ) ] dx [R-r,R+r]
V = 4π ∫ x √ (r² - (x-R)² ) dx [R-r,R+r]

And so then I integrated the whole thing and got 2R - 4Center is (R,0)
r² = (x-R)² + y² ---eqn of a circle at (R,0)
y = ±√ r² - (x-R)²
f(x) = ±√ r² - (x-R)²

h = √ (r² - (x-R)²) - -√( r² - (x-R)²)
h = 2 √ (r² - (x-R)² )

Interval [a,b ] is [R-r,R+r]

V = 2π∫ x h dx [a,b]
V = 2π∫ x [2 √ (r² - (x-R)² ) ] dx [R-r,R+r]
V = 4π ∫ x √ (r² - (x-R)² ) dx [R-r,R+r]

And so I integrated the whole thing and I got 2R - (4π/3)x(-2r^2 - 4Rr)

2. Originally Posted by Reefer
[IMG]file:///C:/Users/c/AppData/Local/Temp/moz-screenshot.png[/IMG]

This problem I do not get at all...

#1 ... method of cylindrical shells

$\displaystyle \displaystyle V = 2\pi \int_a^b r(x) \cdot h(x) \, dx$

$\displaystyle \displaystyle V = 2\pi \int_0^{\sqrt{\frac{\pi}{4}}} x \cdot 12\sin(4x^2) \, dx$

#2 ... Pappus's Centroid Theorem -- from Wolfram MathWorld

2nd Theorem of Pappus (a much easier method in this case) says ...

$\displaystyle V = (2\pi R)(\pi r^2) = 2R(\pi r)^2$

using cylindrical shells ...

let the small circle of radius $\displaystyle r$ be centered at the origin, with rotation radius $\displaystyle (R-x)$.

$\displaystyle \displaystyle V = 2\pi \int_{-r}^r (R-x)(2\sqrt{r^2-x^2}) \, dx$

$\displaystyle \displaystyle V = 4\pi R \int_{-r}^r \sqrt{r^2-x^2} \, dx - 4\pi \int_{-r}^r x\sqrt{r^2-x^2} \, dx$

first integral is just the area of a semicircle of radius $\displaystyle r$ multiplied by the outside constant ... $\displaystyle 2\pi R(\pi r^2)$

using a substitution on the second integral ...

$\displaystyle u = r^2 - x^2$

... note that the limits of integration both become 0 upon substitution.

$\displaystyle \displaystyle V = 4\pi R \int_{-r}^r \sqrt{r^2-x^2} \, dx - 4\pi \int_{-r}^r x\sqrt{r^2-x^2} \, dx = 2R(\pi r)^2 + 0$

... which agrees w/ the calculation using Pappus