Revolution solid

Fantini

MHB Math Helper
Good morning everyone! I have been presented the following problem:

Find the volume of the revolution solid around the $x$ axis of the region between the curves $y=x^2 +1$ and $y=-x^2 +2x +5$ for $0 \leq x \leq 3$.

Finding the intersection of the curves yields $x=-1$ and $x=2$. Therefore, I calculated the integral

$$\pi \int_{-1}^2 [[-x^2 +2x +5]^2 - [x^2 +1]^2] \, dx$$

and found $\frac{189}{3} \pi$. However, the answer is $\frac{277}{3} \pi$. What am I missing? I am also at loss due to the fact that the question proposes the region for $0 \leq x \leq 3$ but it really occurs in $-1 \leq x \leq 2$.

All help is appreciated. Thanks! MarkFL

Staff member
I think what you want to do is:

$\displaystyle V=\pi\left(\int_0^2(-x^2+2x+5)^2-(x^2+1)^2\,dx+\int_2^3 (x^2+1)^2-(-x^2+2x+5)^2\,dx \right)$

This gives you the correct result.

Ackbach

Indicium Physicus
Staff member
You can see from this plot that the two curves only intersect once in the region of interest. You can see by the curvature that the two parabolas will intersect somewhere to the left of $x=0$. You've found that to be at $x=-1$. That's fine, but it's irrelevant. So MarkFL's integral, if you'll notice, is broken up according to where the intersection happens. And you'll also notice that the overall limits of the integral correspond to the region of interest: $[0,3]$. So there you go.

Fantini

Thanks Mark and Ackbach! I gave it some thought about two hours ago, with another person, and I found the answer myself. However, you guys still were a great help! 