# Thread: Finding the volume of this figure

1. ## Finding the volume of this figure

I'm trying to find the volume as shown in the picture I've attached. First of all, would I have to set the functions equal to x rather than y since the figure goes up and down and not left to right, and second of all, I know I'd have to subtract two integrals of something since the figure is hollow in the middle, but I don't know which I'd subtract from which.

2. Originally Posted by fattydq
I'm trying to find the volume as shown in the picture I've attached. First of all, would I have to set the functions equal to x rather than y since the figure goes up and down and not left to right, and second of all, I know I'd have to subtract two integrals of something since the figure is hollow in the middle, but I don't know which I'd subtract from which.
Using the shell method, you can integrate with respect to $x\colon$

$V=2\pi\int_a^bp(x)h(x)\,dx$

$=2\pi\int_0^1(1-x)\left(5\sqrt x-5x^3\right)\,dx$

(here $p(x)$ is the distance from the axis of revolution at $x,$ and $h(x)$ is the height of the region at $x$). This can be easily integrated.

Alternatively, solve for $x$ in the two equations to get $x=y^2/25$ and $x=y^{1/3}/\sqrt[3]5\text.$ Then we have

$V=\pi\int_a^b\left([R(y)]^2-[r(y)]^2\right)dy$

$=\pi\int_0^5\left[\left(1-\frac{y^2}{25}\right)^2-\left(1-\frac{y^{1/3}}{\sqrt[3]5}\right)^2\right]dy\text.$

3. Originally Posted by fattydq
I'm trying to find the volume as shown in the picture I've attached. First of all, would I have to set the functions equal to x rather than y since the figure goes up and down and not left to right, and second of all, I know I'd have to subtract two integrals of something since the figure is hollow in the middle, but I don't know which I'd subtract from which.
hollow? ... are we seeing the correct problem? the directions say to find the volume formed by rotating the region $R_3$ about the line AB, which is x = 1.

4. Originally Posted by skeeter
hollow? ... are we seeing the correct problem? the directions say to find the volume formed by rotating the region $R_3$ about the line AB, which is x = 1.
Definitely hollow. Draw it out.

5. your definition of hollow is different than mine.

to me, hollow means an enclosed empty space.

6. Originally Posted by Reckoner
Using the shell method, you can integrate with respect to $x\colon$

$V=2\pi\int_a^bp(x)h(x)\,dx$

$=2\pi\int_0^1(1-x)\left(5\sqrt x-5x^3\right)\,dx$

(here $p(x)$ is the distance from the axis of revolution at $x,$ and $h(x)$ is the height of the region at $x$). This can be easily integrated.

Alternatively, solve for $x$ in the two equations to get $x=y^2/25$ and $x=y^{1/3}/\sqrt[3]5\text.$ Then we have

$V=\pi\int_a^b\left([R(y)]^2-[r(y)]^2\right)dy$

$=\pi\int_0^5\left[\left(1-\frac{y^2}{25}\right)^2-\left(1-\frac{y^{1/3}}{\sqrt[3]5}\right)^2\right]dy\text.$
And I'm actually supposed to integrate that? Wow...

7. Originally Posted by skeeter
your definition of hollow is different than mine.

to me, hollow means an enclosed empty space.
Right. Exactly. There's empty space that you don't want the volume of.

8. Originally Posted by fattydq
And I'm actually supposed to integrate that? Wow...
First, that should be easy to integrate--just expand. Second, I gave you an easier way to do it using the shell method.

Originally Posted by fattydq
Right. Exactly. There's empty space that you don't want the volume of.
Yes, but it isn't enclosed. Hence the difference in definition.

9. I tried that method. Ended up with 5x^3/2)divided by(3/2)-5x^4/4-2x65/2)+5x^5/5, plugged in 1 5 times on my calculator, multiplied the result by pi every time. same wrong answer every time.

2pi not pi, just realized that...I got it now thanks

10. where did the 1-x come from in your shell method work. That's the only thing i don't understand.

11. Originally Posted by fattydq
where did the 1-x come from in your shell method work. That's the only thing i don't understand.
$1-x$ is the distance between a representative rectangle and the axis of revolution (i.e., the distance between $x$ and the line $x=1$). In other words, it is the average radius of the shell corresponding to each particular value of $x\text.$

Had you been revolving about the $y$-axis, that factor would just be $x;$ for revolution about $x=2,$ it would be (2-x), and so on.

12. Originally Posted by Reckoner
$1-x$ is the distance between a representative rectangle and the axis of revolution (i.e., the distance between $x$ and the line $x=1$). In other words, it is the average radius of the shell corresponding to each particular value of $x\text.$

Had you been revolving about the $y$-axis, that factor would just be $x;$ for revolution about $x=2,$ it would be (2-x), and so on.
I'm sorry could you put that into easier terms? According to my textbook the formula is the integral from a to be of 2pix times f(x)dx