Definite Integrals Free Response - Volume of Solids

[heartshapes]

This is a free response question.. any help would be greatly appreciated!

Let R be the region bounded by the graphs of x=0, y=\(\displaystyle \:4\sqrt{x}\) and y=8
a) Find the volume of the solid generated when R is revolved about the y-axis.

b) Find the volume of the solid generated when R is revolved about the x-axis.

c) The region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a square. Find the volume of the solid.

 

[stapel]

What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you!

Eliz.

 

[heartshapes]

It was actually on my math test I just took.. and I got 4 points out of 9 so I have no idea. I am trying to do test corrections..

for a) I had

\(\displaystyle \pi \int_0^8 {(4\sqrt x )^2 {dy} \]\)
\(\displaystyle \pi\int_0^8 {(16x)} {dy}\) ...
Then my answer was 512\(\displaystyle \[\pi \\).
On this part my teacher circled the y in dy and crossed out the \(\displaystyle \ {4\sqrt x\)

b) it was just a disaster.

c) I had

\(\displaystyle \[\int_0^8 {(8 - 4\sqrt x )^2{\rm }dx}\\)
My answer was 58.56 units[sup:3vriozpu]3[/sup:3vriozpu]. My teacher circled the 8 in the bounds.

Thanks for your help!

 

[stapel]

It looks like you may need to review the formulas. I can't make heads or tails of your set-ups...?

a) Draw the picture. You should have a solid about the y-axis that looks something like a wide-throated vase without any base. (How did you get that 4sqrt[x] = y = 8 intersected at x = 8? The value of 4sqrt[8] is 8sqrt[2], not 8.)

As posted, your integral is with respect to y, and this variable is not included within the integrand, so the integrand will be just a constant. This doesn't make sense in the context, does it? Also, if you are going to integrate using circle-areas, you might want to find an expression which measures the circle radii, not the height, between the curve and the x-axis, of what is not being measured.

b) Draw the picture. You should have a cylinder about the x-axis with a hole dug out of the center; the hole reaches the sides at x = 4, and attenuates, like some sort of thick-sided liquor-glass, to nothing at x = 0.

Since the "slides" here will be "washers", you might want to think about using that formula.

c) Draw the picture. (This will be a bit awkward. Do the best you can.)

Since the cross-sections are all squares, then they must have widths equal to the heights at any point x. For any given x, what is the expression for the height of the shape at that point? (By "height", of course, we mean "top to bottom of the solid slice", not "height of the top of the object, as measured from the table top".)

Given a side-length "s", what is the area A of a square? What will you then need to integrate? Between which x-values?

Eliz.

 

[galactus]

Revolving about the y-axis:

Washers: \(\displaystyle {\pi}\int_{0}^{8}(\frac{y}{4})^{4}dy\)

Shells: \(\displaystyle 2{\pi}\int_{0}^{4}x(8-4\sqrt{x})dx\)


Revolving about the x-axis:

washers: \(\displaystyle {\pi}\int_{0}^{4}(64-16x)dx\)

shells: \(\displaystyle 2{\pi}\int_{0}^{8}y(\frac{y}{4})^{2}dy\)

Part c, the base is \(\displaystyle 8-4\sqrt{x}\), then the cross sections are squares so we have

\(\displaystyle \int_{0}^{4}(8-4\sqrt{x})^{2}dx\)

 

[heartshapes]

okay thank you both for your help!

I fully understand what I did wrong and how to get the right answers!