1. Use cylindrical shells to compute the volume of the solid formed by revolving the region bounded by y= 2x and y = x^2 -3 about x= 6.

2. Find the volume of the solid formed by revolving the region bounded by
y= e^x , y= e^-x, and y= 3 about y= 5.

I'm desperately trying to understand this. I hate volume!

Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.

You got a pair of nasties here.

I will try the first one.

Did you make a sketch?
I got the line and the parabola to intersect at (-1,-2) and (3,6)
since we are rotating about a vertical, the volume has the general appearance of
π[integral] x^2 dy
and we need both equations in term of x
line: x = y/2
parabola : x = ±(y+3)^(1/2)

first consider the region bounded by the line y = 2x and the parabola ABOVE the line from (-1,-2) to (1,-2)
volume of that is ..
π[integral] ((6-y/2)^2 - (6-(y+3)^(1/2))^2 ) dy from -2 to 6
= π[integral] (27 - 7y + y^2/4 - 2(y+3)^(1/2) ) dy
= π { 27y - 7y^2/2 + y^3/12 - (4/3)(y+3)^(3/2) } from -2 to 6
= π(162 - 126 + 18 - 36 - (-54 - 14 - 2/3 - 4/3))
= 88π .... (You better check that arithmetic)

now the part of the parabola from the vertex to the horizontal going through (-1,-2) ....

volume = π[integral] ( (6+(y+3)^(1/2))^2 - (6 - (y+3)^(1/2)^2) dy from -3 to -2
= π[integral] 12(y+3)^(1/2) dy from -3 to -2
= π {16(y+3)^(3/2) } from -3 to -2
= 16π

for a total of 16π + 88π

ok, you better check those calculations. It is so hard to do this kind of math and typing in this kind of setting, and so hard NOT to make some kind of silly error.

try the second one.
Notice you will have symmetry in that one, so use only the part in the first quadrant, then double your answer.

1. Use cylindrical shells to compute the volume of the solid formed by revolving the region bounded by y= 2x and y = x^2 -3 about x= 6.

2. Find the volume of the solid formed by revolving the region bounded by
y= e^x , y= e^-x, and y= 3 about y= 5.

I'm desperately trying to understand this. I hate volume!

First, make a plot of region to be revolved on your calculator. If you do not yet know how to do this, get a friend to show you how. It is a very useful tool. I have a link to a similar plot:
http://img834.imageshack.us/i/1299115553.png/

To calculate using cylindrical shells, you need to cut vertical thin slices of the region and revolve each one around the vertical axis x=6, which coincides with the right side of the above plot.

Let f1(x)=2x, and f2(x)=x^2-3
The intersections of f1(x) and f2(x) are at x=-1 and x=3, or at points (-1,-2), (3,6).

The thin slices have a thickness of dx, at a distance r=6-x from the axis. The height of each slice is therefore h=f1(x)-f2(x). The volume of revolution of each slide is therefore
dV = hdx*2*πr
The total volume, V is
V=∫dV
= ∫2πrhdx from -1 to 3.
Substitute r and h
= ∫2π(6-x)(2x-x^2+3)dx from -1 to 3
Expand the integrand and integrate term by term
= 2π ∫(x^3-8x^2+9x+18)dx
= 2π [(1/4)x^4-(8/3)x^3+(9/2)x^2+18x] [from -1 to 3]
= 320π/3
= 335.1 (approximately)

I will leave #2 for your practice.

Finding the volume of a solid of revolution can be tricky, but I'm here to help you understand the process. Let's break down each problem step by step.

1. Use cylindrical shells to compute the volume of the solid formed by revolving the region bounded by y = 2x and y = x^2 - 3 about x = 6.

To solve this problem using cylindrical shells, we need to consider thin, vertical strips within the region and move them around the axis of rotation. The volume of each strip is calculated as the product of the strip's height, circumference, and thickness. By integrating the volumes of these cylindrical shells over the given interval, we can find the total volume of the solid.

First, let's sketch the region bounded by y = 2x and y = x^2 - 3. We're revolving this region around the vertical line x = 6.

To find the limits of integration, we need to determine where the graphs intersect. Set 2x = x^2 - 3 and solve for x:
x^2 - 2x - 3 = 0
(x - 3)(x + 1) = 0

Therefore, the bounds for integration are x = -1 and x = 3.

Now, for each thin vertical strip, the height (h) is the difference between the two functions: h = 2x - (x^2 - 3) = -x^2 + 2x + 3.

The circumference (C) of each strip is given by the formula C = 2πr, where r is the distance between the axis of rotation (x = 6) and each strip. Since we're revolving around x = 6, r = (6 - x).

The thickness (dx) of each strip is an infinitesimally small change in x.

The volume (dV) of each cylindrical shell is given by dV = 2πr * h * dx.

To find the total volume of the solid, we integrate the volume of each shell over the interval from x = -1 to x = 3:

V = ∫[from -1 to 3] 2π(6 - x)(-x^2 + 2x + 3) dx.

Evaluating this integral will give us the volume of the solid formed by revolving the region around x = 6.

2. Find the volume of the solid formed by revolving the region bounded by y = e^x, y = e^-x, and y = 3 about y = 5.

Here, we have three curves that bound the region we want to rotate around y = 5. To find the volume using cylindrical shells, we use the same basic approach as before.

First, let's sketch the region bounded by y = e^x, y = e^-x, and y = 3. This time, we're revolving the region around the horizontal line y = 5.

To find the limits of integration, we need to determine where the curves intersect. Setting e^x = e^-x and solving for x:

e^x / e^-x = e^2x = 1
x = ln(1) = 0

Therefore, the bounds for integration are x = 0.

For each thin vertical strip, the height (h) is the difference between the top curve (y = 3) and the bottom curve (which varies depending on the x-value).

The circumference (C) of each strip is given by the formula C = 2πr, where r is the distance between the axis of rotation (y = 5) and each strip. Since we're revolving around y = 5, r = (5 - y).

The thickness (dy) of each strip is an infinitesimally small change in y.

The volume (dV) of each cylindrical shell is given by dV = 2πr * h * dy.

To find the total volume of the solid, we integrate the volume of each shell over the interval from y = e^-x to y = e^x:

V = ∫[from e^-x to e^x] 2π(5 - y)(3 - y) dy.

By evaluating this integral, we can determine the volume of the solid formed by revolving the region around y = 5.

Remember that these calculations can be complex, and you might need knowledge of integration techniques to evaluate the integrals.