- $\dfrac{\pi}{2}$
- $\pi$
- $\dfrac{\pi^2}{4}$
- $\dfrac{\pi^2}{2}$
- $\infty$

Solution :

What is the volume of the solid formed by revolving about the x-axis the region in the first quadrant of the xy-plane bounded by the coordinate axes and the graph of the equation: \[y=\frac{1}{\sqrt{1+x^2}}?\]

**Choice 4 is the correct answer.**

Recall that the volume of revolution around the x-axis is given by the formula \[V=\int{\pi y(x)^2}\,dx.\] In this case, we can substitute for $y$ and plug in the limits of integration to obtain: \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx.\]

Almost everyone with some calculus experience has had the great displeasure of memorizing: \[\int{\frac{1}{1+x^2}}\,dx = \arctan x + C.\] Now we can put it to good use! We can see that the answer is (if you'll pardon my poor notation): \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx=\pi\left[\arctan\infty-\arctan0\right]=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}.\]

**Interesting tidbit:** Here's a funny little thing you'll often see in first year calculus. It's called Gabriel's Horn. It's got finite volume but INFINITE surface area. Just something interesting I wanted to share.

- $\dfrac{\pi}{2}$
- $\pi$
- $\dfrac{\pi^2}{4}$
- $\dfrac{\pi^2}{2}$
- $\infty$

Solution :

Recall that the volume of revolution around the x-axis is given by the formula \[V=\int{\pi y(x)^2}\,dx.\] In this case, we can substitute for $y$ and plug in the limits of integration to obtain: \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx.\]

Almost everyone with some calculus experience has had the great displeasure of memorizing: \[\int{\frac{1}{1+x^2}}\,dx = \arctan x + C.\] Now we can put it to good use! We can see that the answer is (if you'll pardon my poor notation): \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx=\pi\left[\arctan\infty-\arctan0\right]=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}.\]

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This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.

There! All dx's and dy's are added. :P

$\[V=\int{\pi y^2}\,dx.\]$

Oops yep. It should be \[V=\int{\pi y(x)^2}\,dx.\]

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