What Solid Is Represented by This Integral?

[STEMucator]

Homework Statement

Interpret the integral as the volume of a certain solid and describe the solid geometrically. Calculate the volume.

$$\int_0^a dy \int_0^{\sqrt{a^2 - y^2}} \frac{2x + 4y}{3} dx$$

Homework Equations

The Attempt at a Solution

Clearly we have an upper surface $z = \frac{1}{3} (2x + 4y)$, which is a plane through the origin since $(x,y) = (0,0) \Rightarrow z = 0$.

The region appears to be described by:

$$R := \{(x,y) \in \mathbb{R^2} \space | \space x=0, y=0, x^2 + y^2 = a^2 \}$$

So we have a cylinder of radius $a$, which we have sliced by the planes $x = 0$ and $y = 0$. This leaves a quarter of a cake piece in the first quadrant.

Then the plane $z = \frac{1}{3} (2x + 4y)$ cuts a chunk out of the bottom half of the cake, creating a quarter cake piece with a slope on its top face aimed towards the origin.

I'm not sure what you would call this solid. Calculating the volume of the integral is easy afterwards.

EDIT: It seems like a sloped quarter disk?

 

[haruspex]

Clearly we have an upper surface $z = \frac{1}{3} (2x + 4y)$, which is a plane through the origin since $(x,y) = (0,0) \Rightarrow z = 0$.

OK, but what is the lower surface?

The region appears to be described by:

$$R := \{(x,y) \in \mathbb{R^2} \space | \space x=0, y=0, x^2 + y^2 = a^2 \}$$

You don't want "=" in there.

 

[STEMucator]

OK, but what is the lower surface?
You don't want "=" in there.

The lower surface is a cylinder. I should have mentioned that. Also, that '=' should read '≤'.

 

[haruspex]

The lower surface is a cylinder.

You're trying to express the integral as the volume of a solid. That means it should correspond to a triple integral ∫∫∫dz.dy.dx over some region. But you are given ∫∫(2x+4y)/3 dy.dx. How do you make that look like a triple integral of "1"?

 

[STEMucator]

You're trying to express the integral as the volume of a solid. That means it should correspond to a triple integral ∫∫∫dz.dy.dx over some region. But you are given ∫∫(2x+4y)/3 dy.dx. How do you make that look like a triple integral of "1"?

In that context we should talk in generality for a moment. Suppose $S$ is the 3-D region in question then the volume of $S$ is given by:

$$V_S = \int \int \int_S dV$$

We know that $S$ is the region below $z = \frac{1}{3} (2x + 4y)$ and above the region $R$ in the x-y plane. So we can re-write the integral:

$$V_S = \int \int \int_S dV = \int \int_R \left[\int_0^z dz \right]dA = \int \int_R z dA = \int \int_R \frac{1}{3} (2x + 4y) dA$$

EDIT: Just to be a little more rigorous, I should mention:

$$S := \{ (x,y,z) \space | \space (x,y) \in R, \space x=0, \space y = 0, \space x^2 + y^2 \leq a^2, \space 0 \leq z \leq \frac{1}{3} (2x + 4y) \}$$

 

[haruspex]

$$S := \{ (x,y,z) \space | \space (x,y) \in R, \space x=0, \space y = 0, \space x^2 + y^2 \leq a^2, \space 0 \leq z \leq \frac{1}{3} (2x + 4y) \}$$

Right, this is the lower surface: 0 ≤ z.
Note, though, that this is arbitrary. You could have validly chosen 1 ≤ z ≤ (2x + 4y)/3 + 1, etc.