Finding Volume in Cylinder

[trumpet-205]

Homework Statement

The problem is in Calculus Early Transcendentals by Stewart Ed.6th on page 448 problem #4.

The question is about a cylinder that is tilted with a radius $$r$$ and a height $$L$$ is filled with water that cover the lower base and touch the edge of upper base. It asked to set it up in two ways to find its volume. First to slice it parallel in rectangular section. Second to slice it parallel in trapezoids section. I don't know how to set it up. ?

I am also having trobule in part e, where the water is now covering half of the lower base but it still touches the edge of upper base. Now it asked me to slice it in triangle, rectangles, and segments of circles.

Homework Equations

Cylinder is tilted with radius of r and height of L.
Volume of cylinder - $$\pi r^{2}L$$

An equation my instructor gives - $$x^{2} + y^{2} = r^{2}$$

$$0 \leq z \leq L$$

The Attempt at a Solution

No idea...

 

[mighty2000]

Thank you for reaching out for help with this problem. I can understand your confusion as it can be challenging to visualize and set up a problem involving a tilted cylinder. However, I am here to assist you in understanding and solving this problem.

To begin, let's consider the first part of the problem where the cylinder is sliced parallel in rectangular sections. In this case, we can imagine the cylinder being sliced into thin horizontal layers, each with a height of Δz. The first layer will have a base with a length of 2r and a width of Δz, as shown in the attached diagram.

[Insert diagram of cylinder sliced into rectangular sections]

We can now calculate the volume of this first layer using the formula for the volume of a rectangular prism, V = lwh. In this case, the length is 2r, the width is Δz, and the height is Δz. Therefore, the volume of this first layer is V = 2rΔz^2.

We can continue this process for each layer, each with a base of 2r and a height of Δz, until we reach the top of the cylinder at z = L. To find the total volume, we can add up the volumes of each layer using integration. The integral for the volume of the entire cylinder would be:

∫V dz = ∫2rΔz^2 dz, from z = 0 to z = L

Solving this integral will give us the volume of the cylinder.

Now, for the second part of the problem where the cylinder is sliced parallel in trapezoidal sections, we can imagine the cylinder being sliced into thin horizontal layers again. However, this time, the base of each layer will be a trapezoid with a larger base of 2r and a smaller base of 2r - 2Δz, as shown in the attached diagram.

[Insert diagram of cylinder sliced into trapezoidal sections]

Using the formula for the area of a trapezoid, A = ½h(b1 + b2), we can calculate the area of each layer. The height of each layer is still Δz, and the two bases are 2r and 2r - 2Δz. Therefore, the volume of each layer would be V = ½Δz(2r + 2r - 2Δz)