Calculating the Volume of a Cylindrical Wedge Using Calculus and Geometry

[RK7]

Homework Statement

Find the volume of the enclosed by the surfaces $$z=qx$$ $$z=0$$ and $$x²+y²=2ax$$

Homework Equations

This is meant to be done with calculus but can verify my answer with simple geometry - should be $$\pi a^3q$$

The Attempt at a Solution

So the top of the wedge will be when $$x=2a$$

Form rectangular slices of the wedge perpendicular to the x-axis with area $$A=2yz=2\sqrt{x(2a-x)}qx$$ and volume $$2\sqrt{x(2a-x)}qx .dx$$

Then integrate this from x=0 to x=2a gives:
V=$\int ^{2a} _{0} 2q \sqrt{x^{3}(2a-x)} dx$


I've checked this numerically with wolfram alpha for certain values of a and q but I haven't got a clue how to evaluate it.. the question said to use a double integral but I don't know what a suitable double integral would be...

 

[RK7]

Hm I've come up with a double integral:

Break it up into segments of the cirlce in the plane perpendicular to z axis

The area of a segment at height z is $$\int^{2a} _{z/q} 2y dx = \int^{2a} _{z/q} 2 \sqrt{x(2a-x)} dx$$ and can integrate these from z=0 to 2aq giving:
$$\int^{2aq} _{0} \int^{2a} _{z/q} 2 \sqrt{x(2a-x)} dx dz$$

 

[RK7]

Similar problem with evaluating the integral though..

 

[DryRun]

Are you sure the answer isn't $$\frac {\pi a^3q}{2}?$$

 

[RK7]

Are you sure the answer isn't $$\frac {\pi a^3q}{2}$$?

So the cylinder has radius a and touches x=0 and x=2a
The cylinder goes up to a height of z=qx=2aq
So the total volume of the cylinder is $$\pi a^{2} * 2aq = 2 \pi a^{3} q$$
So the volume of the half of the cylinder we're interested in is half that $$V=\pi a^3 q$$

Am I doing something wrong?

 

[DryRun]

What you just calculated in #5 is the volume of a normal vertically-halved cylinder of height z=2aq (upper surface being the horizontal plane z=2aq), when you should be calculating a cylindrical wedge.

Let q=0. What does this tell you about the volume?

 

[RK7]

What you just calculated in #5 is the volume of a normal vertically-halved cylinder of height z=2aq (upper surface being the horizontal plane z=2aq), when you should be calculating a cylindrical wedge.

Let q=0. What does this tell you about the volume?

What's wrong with the answer $$\pi a^3 q$$ ?

And q=0 is just zero volume..

 

[DryRun]

You need to divide your answer by 2 again. Why?

1. You are dealing with half of a cylinder (vertically split in half).
2. You are dealing with a wedge in the cylinder (diagonally split in half).

 

[RK7]

You need to divide your answer by 2 again. Why?

1. You are dealing with half of a cylinder (vertically split in half).

I'm dealing with an entire cylinder split in two. The equation is x²+y²=2ax (equivalent ton (x-a)²+y²=a²) not x²+y²=a².

 

[DryRun]

I'm dealing with an entire cylinder split in two. The equation is x²+y²=2ax (equivalent ton (x-a)²+y²=a²) not x²+y²=a².

That's what i meant. Enter all the equations in a 3D software to see the volume required.

 

[RK7]

Can you upload a screenshot of what your software is showing?

 

[DryRun]

I have drawn it on paper as you would in the exams. I suggested using a graphing software to help you to visualize it in 3D.

Here is the double integral (using polar coordinates):
$$\int^{\theta = \frac{\pi}{2}}_{\theta = 0} \int^{r=2a\cos \theta}_{r=0} \left[ z \right]^{qr\cos \theta}_0\,.rdrd\theta$$

 

[RK7]

I have drawn it on paper as you would in the exams. I suggested using a graphing software to help you to visualize it in 3D.

Here is the double integral (using polar coordinates):
$$\int^{\theta = \frac{\pi}{2}}_{\theta = 0} \int^{r=2a\cos \theta}_{r=0} \left[ z \right]^{qr\cos \theta}_0\,.rdrd\theta$$

With respect, I'm fairly certain you're wrong and completely derailed the thread.