Solving Homework: Polar Coordinates Issue on Volume

[STEMucator]

Homework Statement

My answer seems to differ from the books answer, so I'm wondering where something has gone wrong.

Find the volume inside the prism bounded by the planes $y = x$, $y = 0$, $x = \frac{a}{\sqrt{2}}$, $z = 0$ and the cone $az = h\sqrt{x^2 + y^2}$.

Homework Equations

We want the area under $z = \frac{h}{a} \sqrt{x^2 + y^2}$

The Attempt at a Solution

When i visualized this, I saw a triangle in the x-y plane formed by $y = x$, $y = 0$, and $x = \frac{a}{\sqrt{2}}$, which gets cut by $z = \frac{h}{a} \sqrt{x^2 + y^2}$ in the x-y-z plane.

So we want:

$$V_R = \int \int_R \frac{h}{a} \sqrt{x^2 + y^2} dA$$

Using polar co-ordinates gives:

$(1) rsin(\theta) = 0 \quad$
$(2) rsin(\theta) = rcos(\theta)$
$(3) rcos(\theta) = \frac{a}{\sqrt{2}}$
$(4) f(rcos(\theta), rsin(\theta)) = \frac{h}{a} r$

Looking at $(2)$ implies $\theta = \frac{\pi}{4}$ so that $0 ≤ \theta ≤ \frac{\pi}{4}$.

The limits for $r$ have proven a little tricky. The answer is listed as:

$\frac{a^2 h}{12} [1 + \frac{3 \sqrt{2}}{2} \log(1 + \sqrt{2})]$

I have obtained the answer $\frac{a^2 h \pi}{12}$ using $0 ≤ r ≤ a$, which is obviously incorrect, but indicates my limits for $\theta$ make physical sense, so I believe the limits for $r$ are the issue. I can't seem to visualize them properly right now.

 

[jedishrfu]

...deleted... My answer was off...

 

[STEMucator]

Do you need to use calculus?

I drew a picture of it and it looks like cylinder with a cone inside and y=x and y=0 planes indicate 1/8 of the volume.

You should already know the volume of a cylinder and a cone right.

Calculus is the bread and butter of all that is true. So forgive me if I'm firm about using it.

Where is the cylinder coming from exactly? The lines $y=0$, $y=x$ and $x = \frac{a}{\sqrt{2}}$ give a triangle with a varying height in the x-y plane (based on $\frac{a}{\sqrt{2}}$).

 

[jedishrfu]

Calculus is the bread and butter of all that is true. So forgive me if I'm firm about using it.

Where is the cylinder coming from exactly? The lines $y=0$, $y=x$ and $x = \frac{a}{\sqrt{2}}$ give a triangle with a varying height in the x-y plane (based on $\frac{a}{\sqrt{2}}$).

I thought I saw a cylinder with cone defining its upper edge but had forgotten about the x=a/sqrt(2) plane

 

[vela]

Why are you even using polar coordinates here? It's more straightforward to describe the area of integration using cartesian coordinates.

 

[STEMucator]

Re-arranging the equation for the cone a bit:

$az = h \sqrt{x^2 + y^2} \Rightarrow \frac{z^2}{h^2} = \frac{x^2}{a^2} + \frac{y^2}{a^2}$

I see that it opens up in the z-direction and the top of the cone should be a perfect circle of radius $a$. I still can't see how to get the radius otherwise.

The book says it is possible to find the solution with polar co-ordinates, and I would like to find a solution using them.

 

[LCKurtz]

You have that triangle you described in the xy plane. Use it to get the polar limits. $r$ goes from $0$ to the $r$ value on the line $x=\frac a{\sqrt 2}$. What is the equation of that line in polar coordinates?

 

[STEMucator]

You have that triangle you described in the xy plane. Use it to get the polar limits. $r$ goes from $0$ to the $r$ value on the line $x=a\sqrt 2$. What is the equation of that line in polar coordinates?

That would correspond to equation $(3)$ in my first post I believe.

$$r \cos(\theta) = \frac{a}{\sqrt{2}}$$

Though I doubt that $0 ≤ r ≤ \frac{a}{cos(\theta) \sqrt{2}}$

 

[LCKurtz]

That would correspond to equation $(3)$ in my first post I believe.

$$r \cos(\theta) = \frac{a}{\sqrt{2}}$$

Though I doubt that $0 ≤ r ≤ \frac{a}{cos(\theta) \sqrt{2}}$

Why do you doubt it? If you start at $r=0$ and move through the region in the $r$ direction, isn't that the line you hit? And I would suggest expressing that with $\sec\theta$ when you set up the integral.

 

[STEMucator]

Why do you doubt it? If you start at $r=0$ and move through the region in the $r$ direction, isn't that the line you hit? And I would suggest expressing that with $\sec\theta$ when you set up the integral.

I am never going to believe wolfram again.

I think I understand now that you have clarified that. Thank you.