# Decide volume given two functions

• MHB
• Nivelo
In summary, the conversation discusses calculating the volume of a solid created when an area in the first quadrant is rotated around the y-axis. The speaker suggests using the disc method, but runs into issues due to one of the functions crossing the x-axis. They then recommend using the cylinder method instead, breaking the solid into two parts and calculating the volume using integrals. The final answer is determined to be 52*pi/6.
Nivelo
Sorry if i made any language errors, english is not my first language.
Question: An area in the first quadrant (x=>0,y=>0) is limited by the axis and the graphs to the functions f(x)=x^2-2 and g(x)=2+x^2/4. When the area rotates around the y-axis a solid is created. Calculate the volume of this solid.

I want to calculate with the disc method. I set h(x)=g(x)-f(x) since g(x)>f(x) in this area. Since x>=0 i check where the lower function crosses the x-axis and get
x^2-2=0 which gives x=sqrt(2) (since x can't be negative). The upper limit is where f(x)=g(x), so
x^2-2=x^2/4+2 which gives x=4/sqrt(3).
One of the discs that i want to sum up should have the volume pi*radius^2*dy.I get the radius from solving x^2 from h(x)=y.And so i get that the volume should be
I=pi/3* integral from 0 to 10/3 of (16-4*y)dy. But its not the right answer and i don't know where i went wrong. Could anyone tell me where i messed up and try to help me solve this one?

Thanks in advance, appreciate all the help!

You are going to have a problem with the "disk method"! From y= -2 to y= 2, your "disks" are entirely bounded by the first function $$y= x^2- 2$$ so, for a given y, the radius is $$x= \sqrt{y+ 2}$$. However, from y= 2 to $$\frac{4}{\sqrt{2}}= 2\sqrt{2}$$ your disk has a "hole" in it because the second parabola, $$y= 2+ \frac{x^2}{4}$$ comes into play. You could use the disk method from y= -2 to y= 2 and then use the "washer method" from y= 2 to $$y= 2\sqrt{2}$$.

But I would recommend, instead, the "cylinder method", using x as the variable, rather than y. For a given x, we have a cylinder with inner radius x, height from $$y= x^2- 2$$ up to $$y= 2+ \frac{x^2}{4}$$ so height is $$2- \frac{x^2}{4}- (x^2- 2)= 4- \frac{5x^2}{4}$$. That is, the circumference is $$2\pi x$$, the surface area is $$2\pi x\left(4- \frac{5x^2}{4}\right)= 2\pi\left(4x- \frac{5x^3}{4}\right)$$ and, with thicknes dx, volume $$2\pi\left(4x- \frac{5x^3}{4}\right)dx$$. Integrate that from x= 0 to $$x= \frac{4\sqrt{3}}{3}$$.

disks & washers about the y-axis ...

$$\displaystyle V = \pi \int_0^2 (y+2) \, dy + \pi \int_2^{4/\sqrt{3}} (y+2) - 4(y-2) \, dy$$

note ... images should be r(y) for the disks and R(y) & r(y) for the washers

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HallsofIvy said:
You are going to have a problem with the "disk method"! From y= -2 to y= 2, your "disks" are entirely bounded by the first function $$y= x^2- 2$$ so, for a given y, the radius is $$x= \sqrt{y+ 2}$$. However, from y= 2 to $$\frac{4}{\sqrt{2}}= 2\sqrt{2}$$ your disk has a "hole" in it because the second parabola, $$y= 2+ \frac{x^2}{4}$$ comes into play. You could use the disk method from y= -2 to y= 2 and then use the "washer method" from y= 2 to $$y= 2\sqrt{2}$$.

But I would recommend, instead, the "cylinder method", using x as the variable, rather than y. For a given x, we have a cylinder with inner radius x, height from $$y= x^2- 2$$ up to $$y= 2+ \frac{x^2}{4}$$ so height is $$2- \frac{x^2}{4}- (x^2- 2)= 4- \frac{5x^2}{4}$$. That is, the circumference is $$2\pi x$$, the surface area is $$2\pi x\left(4- \frac{5x^2}{4}\right)= 2\pi\left(4x- \frac{5x^3}{4}\right)$$ and, with thicknes dx, volume $$2\pi\left(4x- \frac{5x^3}{4}\right)dx$$. Integrate that from x= 0 to $$x= \frac{4\sqrt{3}}{3}$$.

Thanks for the response, i solved it. Since one of the demands was that y>=0 i broke the solid down into two parts, one from x=0 to x=sqrt(2) which is before the lower function crossed the x-axis f(x)=0. The other part was from where the lower function crosses the x-axis f(x)=0 to where the upper and lower function crosses f(x=g(x) ,which gives x=4/sqrt(3). From x=0 to x=sqrt(2) the height of one "rectangle" is the distance from the upper function to the y-axis and from x=sqrt(2) to x=4/sqrt(3) the height is the difference g(x)-f(x).
Volume = (2*pi*integral from 0 to sqrt(2) of g(x)) + (2*pi*integral from sqrt(2) to 4/sqrt(3) of g(x)-f(x)) = 52*pi/6

## 1. What is the purpose of determining volume given two functions?

The purpose of determining volume given two functions is to find the volume of a three-dimensional shape that is formed by the intersection of two functions. This can be useful in various fields such as engineering, physics, and architecture.

## 2. How do you calculate the volume given two functions?

To calculate the volume given two functions, you first need to identify the boundaries of the shape by finding the points of intersection between the two functions. Then, you can use the formula for volume of a solid of revolution, which involves integrating the difference between the two functions over the identified boundaries.

## 3. Can the volume given two functions be negative?

No, the volume given two functions cannot be negative. Volume is a measure of space and therefore, it is always a positive value. If the calculated volume is negative, it indicates an error in the calculation.

## 4. What is the difference between volume given two functions and volume given one function?

The main difference between volume given two functions and volume given one function is that in the former, the shape is formed by the intersection of two functions while in the latter, the shape is formed by rotating a single function around an axis. The calculations for both types of volumes also differ.

## 5. Are there any real-world applications of determining volume given two functions?

Yes, there are many real-world applications of determining volume given two functions. For example, in engineering, this concept is used to calculate the volume of a 3D object such as a pipe or a tank. In architecture, it can be used to determine the volume of a building or a room. In physics, it can be used to calculate the volume of a solid object such as a planet or a star.

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