Volume of Revolution

**Dickson** · June 8th 2009, 07:11 AM

Find the volume of revolution when the area between $y=x^2 + 1$ and $y=x+3$ is rotated about the x-axis.

$(x+3)^2 = x^2 + 6x + 9$
$(x^2+1)^2 = x^4 +2x^2 + 1$

So subtract those and I got:

$-x^4 - x^2 + 6x + 8$

Using S for integral as I don't know the code for it.

$V = (pie)S(-1 to 5) (-x^4 - x^2 + 6x + 8)$
Going to jump into the substitution with the anti-deriv.
$=(pie)(-(2)^5 / 5 - (2)^3 / 3 + 3(2)^2 + 8(2)) - (-(-1)^5 / 5 - (-1)^3 / 3 + 3(-1)^2 + 8(-1))$

Got a final answer of $119/5$ but my friend got a different answer of $77/5$ . What did I do wrong if mine is incorrect

**Random Variable** · June 8th 2009, 07:48 AM

The limits are wrong. It's from -1 to 2.

**Dickson** · June 8th 2009, 07:51 AM

Originally Posted by Random Variable

The limits are wrong. It's from -1 to 2.

Ooops wrote that wrong...the work on my paper uses 2

**Random Variable** · June 8th 2009, 08:19 AM

I'm getting $\frac {117}{5} \pi$

$\pi \int^{2}_{-1} \Big( (x+3)^2 - (x^{2}+1) \big) \ dx$

$= \pi \int^{2}_{-1} \Big(-x^{4}-x^{2}+6x+8 \Big) \ dx$

$= \pi \Big( -\frac {1}{5}x^{5} - \frac {1}{3} x^{3} + 3x^{2} + 8x \Big) \Big|^{2}_{-1}$

$\pi \bigg(\Big( -\frac {32}{5} - \frac {8}{3} + 12 + 16 \Big) - \Big( \frac {1}{5} + \frac {1}{3} + 3 - 8 \Big) \bigg)$

$= \frac {117}{5} \pi$

**calc101** · June 8th 2009, 08:21 AM

The radii are incorrect. You are rotating the region around the x-axis. The closest point to the x-axis of the function $x^2+1$ is (0,1) and the closest point of the function $x+3$ is (-1,2). Therefore, the radii should be:

$Rin = (x^2 + 2)^2$
$Rout = (x + 5)^2$

So, your integral should be something like this:

$V = \pi \int -x^4-3x^2+10x+21 dx$ , from -1 to 2

I got: $\frac{312\pi}{5}$

Edit: Can someone verify my answer, please.

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