# Thread: finding volume of a shape

1. ## finding volume of a shape

I need to find the volume of a solid.
between the curve y = sec^(-1)(x) and the x-axis from x=1 and x=2 revolved around the y-axis.
using both the slicing method and cylindrical shell method.
sec^(-1) is sec inverse.

volumes of normal shapes i can find, but i think the sec is throwing me off.

any help along the right path would be appreciated.
thanks

2. Originally Posted by rudy
I need to find the volume of a solid.
between the curve y = sec^(-1)(x) and the x-axis from x=1 and x=2 revolved around the y-axis.
using both the slicing method and cylindrical shell method.
sec^(-1) is sec inverse.

volumes of normal shapes i can find, but i think the sec is throwing me off.

any help along the right path would be appreciated.
thanks
If we are to use the slicing method, what would be the shape of the cross-section that is perpendicular to this region?

For cylindrical shells, we know that if an object is rotated about y, we see that $V=2\pi\int_1^2 x\sec^{-1}x\,dx$

To integrate this, apply integration by parts: let $u=sec^{-1}x$ and $\,dv=x\,dx$

However, keep this in mind as you evaluate the integral:

$\frac{1}{|x|\sqrt{x^2-1}}=\left\{\begin{array}{rl}\displaystyle-\frac{1}{x\sqrt{x^2-1}}&x<-1\\\displaystyle\frac{1}{x\sqrt{x^2-1}}&x> 1\end{array}\right.$

Can you take it from here?

--Chris

3. ok, cylindrical is making a tad more sense. i did the integral and got a value of 7.69, does this sound right to you?

and for the slicing method, i would assume to use a disk or circle shape. Area would be pi(R^2-r^2), where r would be x. but i have no idea how to put y=sec^(-1)(x) in terms of x.
does this sound right or have i completely lost it? haha

4. Originally Posted by rudy
ok, cylindrical is making a tad more sense. i did the integral and got a value of 7.69, does this sound right to you?

and for the slicing method, i would assume to use a disk or circle shape. Area would be pi(R^2-r^2), where r would be x. but i have no idea how to put y=sec^(-1)(x) in terms of x.
does this sound right or have i completely lost it? haha
I see. You are trying to find the volume using washers

I would recommend getting the integral in terms of y, instead of x.

So, $y=\sec^{-1}x\implies x=\sec y$. Can you prove that our integral would be $\pi\int_0^{\frac{\pi}{3}}\left[4-\sec^2y\right]\,dy$?? [hint: plot $y=\sec^{-1}x$ over the interval [1,2], and then make the adjustments from x to y]

I'll let you work on showing this [and then evaluating]. If you get stuck, let us know.

--Chris

5. ive tried and tried but just can get that integral.

ive solved the integral and got the right answer, corresponded with the cylindrical method.

6. Originally Posted by rudy
ive tried and tried but just can get that integral.

ive solved the integral and got the right answer, corresponded with the cylindrical method.
This is how I would set up the volume integral, using the washer method:

Here is a graph of $y=\sec^{-1}x$ over the interval [0,2] [I highlighted the region we will be using]:

To apply washer method:

If region is rotated about y axis: $V=\pi\int_c^d \left[(g_2(y))^2-(g_1(y))^2\right]\,dy$

If region is rotated about x axis $V=\pi\int_a^b \left[(f(x))^2-(g(x))^2\right]\,dx$

In our case, its being revolved about the y axis.

So we need to get the function in terms of $x=f(y)$

So we see that $y=\sec^{-1}x\implies x=\sec y$

I will re-plot the graph in terms of y:

In this graph, the horizontal is the y axis, and the vertical is the x axis.

So in this graph, we will be rotation this region about the horizontal [y axis]

In this case, our $g_2(y)=2$ and our $g_1(y)=\sec y$, and the limits on y will be from 0 to $\frac{\pi}{3}$ [I leave it for you to show why these are the proper limits]

So our volume integral would then be $V=$ $\pi\int_c^d \left[(g_2(y))^2-(g_1(y))^2\right]\,dy=\pi\int_0^{\frac{\pi}{3}}\left[(2)^2-(\sec y)^2\right]\,dy=\pi\int_0^{\frac{\pi}{3}} \left(4-\sec^2 y\right)\,dy$

Is it clearer now on how to set up this volume integral?

--Chris

7. aaaaahhh yes of course.

thanks very very much.
seems so simple now that you know the answer.