|
This question stumped me on a recent calculus test... I would really like an explanation. The region $\mathbf{R}$ is bounded by $h(x) = -0.8(x-2)^2 + 8$ and $p(x) = x\cos(x) - \ln(2x+2)$. Find the lateral surface area of the solid formed when $\mathbf{R}$ is revolved around the line $y=11$. I tried to convert both functions into terms of y so that I could get the arc length, but that didn't get me very far. Any help would be greatly appreciated. |
|||||
|
|
First, let's shift the functions down so that instead of revolving about $y=11$, we're revolving about the x-axis, $y=0$: $h_1(x)=h(x)-11$, $p_1(x)=p(x)-11$. Let $a< b$ be the x-coordinates of the points of intersection of $h_1(x)$ and $p_1(x)$ (in the vicinity of -2 and 5)--that is, the least and greatest x-coordinates of any points in $\mathbf{R}$. There are two lateral surface areas to compute: the "inner" part bounded by $h_1$, and the "outer" part bounded by $p_1$. For each, the surface area is given by $\int_{a}^{b}2\pi y(x)\sqrt{1+(y'(x))^2}dx$, replacing $y$ with $h_1$ or $p_1$ (and $y'$ with its derivative) as appropriate. |
|||
|
