# Calculate Volume of Rotated Geometry: y-10=x and y2 -6y =x, around x=1

• Lo.Lee.Ta.
In summary, the conversation is about finding the volume between two curves and rotating it around a given axis. The correct answer is 75.6*pi, but there was some confusion about the calculations and inputting the equation into Wolfram Alpha. The mistake was found and corrected, and it was noted that using pi(x) in Wolfram Alpha can lead to misunderstandings.
Lo.Lee.Ta.
1. Find the volume between y-10=x and y2 -6y =x, rotated around x=1.

2. R= y2 -6y -1

r= y-10 -1

∫2 to 5 of [($\pi$(y2 -6y -1)2) - $\pi$(y - 11)2]

∫$\pi$[y4 - 12y3 + 4y2 + 12y + 1] - $\pi$[y2 - 22y + 121]dy

= $\pi$[y5/5 - 12y4/4 + 4y3/3 +122/2 +y] - $\pi$[y3/3 - 22y2/2 + 121y] |2 to 5

...LONG SUBSTITUTION...

= -928.33$\pi$ - 371.67$\pi$ + 4.93$\pi$ + 200.67$\pi$

= -1094.4$\pi$

...Is this the right answer...? I don't really think it's right because it's negative!
But what am I doing wrong here? :(

Thanks SO much for your help! :D

Lo.Lee.Ta. said:
1. Find the volume between y-10=x and y2 -6y =x, rotated around x=1.2. R= y2 -6y -1

r= y-10 -1∫2 to 5 of [($\pi$(y2 -6y -1)2) - $\pi$(y - 11)2]

∫$\pi$[y4 - 12y3 + 4y2 + 12y + 1] - $\pi$[y2 - 22y + 121]dy

= $\pi$[y5/5 - 12y4/4 + 4y3/3 +122/2 +y] - $\pi$[y3/3 - 22y2/2 + 121y] |2 to 5

...LONG SUBSTITUTION...

= -928.33$\pi$ - 371.67$\pi$ + 4.93$\pi$ + 200.67$\pi$

= -1094.4$\pi$

...Is this the right answer...? I don't really think it's right because it's negative!
But what am I doing wrong here? :(

Thanks SO much for your help! :D

It's easy to get something negative if you are subtracting something larger from something smaller when you meant to do the reverse. Which of those curves has the largest radius around the axis x=1? And I don't think just flipping the sign right will fix it either. Some other error around as well. Ah, you didn't square y^2-6y-1 correctly.

You're right! I didn't square the (y2 - 6y -1) correctly! X(

The correct sqaring should equal: (y4 - 12y3 + 34y2 + 12y + 1)

So the integral seems like is should really be:

∫2 to 5 [$\pi$(y4 - 12y3 + 34y2 + 12y + 1) - $\pi$(y2 - 22y + 121)]dy

AWFUL AND LONG SUBSTITUTION!

= 321.67$\pi$ - 371.67$\pi$ - 75.07$\pi$ + 199.33$\pi$

= 74.26$\pi$

BUT when I put this whole integral into Wolfram Alpha (including the squares), it said it equals 0!

Then when I put in the integral again (this time with the squares factored out), it said the answer should be 15.9016!
What?! Would you please tell me what's wrong here?

Thank you SO much! :D

I went through this thing again and again, and I only thing I found was a slight error making the answer 75.6$\pi$!

Lo.Lee.Ta. said:
I went through this thing again and again, and I only thing I found was a slight error making the answer 75.6$\pi$!

I think 75.6*pi sounds just fine. You KNOW 0 isn't correct, right? What exactly did you input to WA?

integral from 2 to 5 of [pi(y^4 - 12y^3 + 34y^2 + 12y + 1) - pi(y^2 - 22y + 121)]

The above is exactly what I typed into WA.

Lo.Lee.Ta. said:
integral from 2 to 5 of [pi(y^4 - 12y^3 + 34y^2 + 12y + 1) - pi(y^2 - 22y + 121)]The above is exactly what I typed into WA.

That's an interesting misunderstanding. WA takes pi(x) to be a function giving the number of primes less than or equal to x. http://en.wikipedia.org/wiki/Prime-counting_function Try reading the fine print in the input box. If don't want that write pi*(x) instead.

Last edited:

## 1. How do you calculate the volume of rotated geometry?

To calculate the volume of rotated geometry, you need to use the formula V = π∫(R(x))^2dx, where R(x) is the function describing the shape being rotated and the integral is taken over the interval of rotation.

## 2. What is the meaning of "y-10=x" and "y2 -6y =x" in the context of this calculation?

The equations y-10=x and y2 -6y =x represent two intersecting curves that define the boundaries of the rotated shape. These equations are used to determine the limits of integration in the volume calculation.

## 3. What is the significance of x=1 in this calculation?

The value of x=1 represents the axis of rotation. This means that the shape is being rotated around a vertical line passing through the point (1,0).

## 4. Can this calculation be done without using calculus?

No, this calculation requires the use of calculus (specifically integration) to find the volume of the rotated geometry.

## 5. How can this calculation be applied in real life situations?

The calculation of volume of rotated geometry can be applied in various engineering and architectural fields, such as designing curved structures, calculating the volume of rotating objects, and determining the capacity of containers with curved walls.

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