MHB Find the volume by revolving around the x-axis without expanding

[karush]

A solid is generated when the region in the first quadrant enclosed by the graph of , $$y=(x^2+1)^2$$
The line $$x=1$$ , the x-axis, and the y-axis is revolved about the x-axis. Find the volumn

$$
V=\int_{a}^{b}\pi\left[f(x)\right]^2 dx
=\int_{0}^{1}\pi\left[\left(x^2+1\right)^2\right]^2 \,dx
=\int_{0}^{1}\pi\left(x^2+1\right)^4 \,dx
$$

provided this is ok so far, how do you do this without expanding it?

 

[I like Serena]

A solid is generated when the region in the first quadrant enclosed by the graph of , $$y=(x^2+1)^2$$
The line $$x=1$$ , the x-axis, and the y-axis is revolved about the x-axis. Find the volumn

$$
V=\int_{a}^{b}\pi\left[f(x)\right]^2 dx
=\int_{0}^{1}\pi\left[\left(x^2+1\right)^2\right]^2 \,dx
=\int_{0}^{1}\pi\left(x^2+1\right)^4 \,dx
$$

provided this is ok so far, how do you do this without expanding it?

Hi karush,You can do it by repeated partial integration:
$$\int udv = uv - \int vdu$$
So:
$$\begin{aligned}\int_0^1 (x^2+1)^4 dx
&= (x^2+1)^4 x \Big|_0^1 - \int_0^1 x\cdot d((x^2+1)^4)\\
&= 16 - \int_0^1 x\cdot 4(x^2+1)^3 \cdot 2x\,dx \\
&= 16 - 8\int_0^1 (x^2+1)^3d(\frac 1 3 x^3)
\end{aligned}$$
and repeat...... but it's easier to expand and integrate! ;)

 

[Dethrone]

Hi Karush (Wave),

Personally, I don't see a way around it...I could be wrong. Solving it as a function of $y$ seems to be even more difficult.

 

[karush]

ok expanding looks easier so

expanding $\left(x^2+1\right)^4$ would be $x^8+4x^6+x^4+4x^2+1$

so then we have

$$\pi\int_{0}^{1}\left(x^8+4x^6+6x^4+4x^2+1\right) dx$$

 

[I like Serena]

Your coefficients aren't quite right...

 

[karush]

is post #4 correct I edited it..(Dull)

 

[Dethrone]

Yes, it is. (Yes)

 

[I like Serena]

is post #4 correct I edited it..(Dull)

Yes! Your integral is correct now.
(But your expansion still has the wrong coefficient.)