Find the volume of the solid formed by rotating the region enclosed by
y=e^(5x) , \ y=0, \ x=0, \ x=0.8
around the y-axis
Please help, i have been attempting these problems for a couple of days
the y-intercept of y = e^(5x) is (0,1)
and y = e^(5x) intersects x = 0.8 at (.8 , 54.598)
so I made a sketch , and I am looking at a region made up of a rectangle topped by a another region resembling a triangle.
the volume of the cylinder with radius .8 and height 54.598 is
V = π(.8)^2 ( 54.598) = appr. 109.776
we now have to hollow-out the part which is not part of the solid, and we have to solve y = e^(5x) for x
y = e^(5x)
ln y = 5x
x = (1/5)lny
so the volume of the triangular - looking region
= π∫(1/25) (lny)^2 dy from 1 to 54.598
At this point I must admit that I "cheated" and used Wolfram to find the integral
http://integrals.wolfram.com/index.jsp?expr=%28ln%28x%29%29%5E2&random=false
= (1/25)π [ y(ln y)^2 - 2y ln y + 2y ] from 1 to 54.598
= ...
I got appr. 68.358
so that would give us a volume of 109.776 - 68.358 = 41.42
You better check my arithmetic, I would suggest you differentiate my integral to make sure you get (1/25) (ln y)^2
(just do the y(ln y)^2 - 2y ln y + 2y part)
or, if you like to use shells, you have
v = integral [0,0.8] 2πrh dx
where r = x h = y = e^(5x)
v = 2π integral[0,0.8] xe^)5x) dx
= 1/25 e^(5x) (5x-1)
= 41.417
Hmmm. using washers, I get
v = π*.64*1 + integral[1,e^4] π (R^2-r^2) dy
where R = .8 and r = 1/5 ln y
v = .64π + π integral[1,e^4] (.64 - 1/25 ln^2(y)) dy
= .64π + 12.544π = 13.184π = 41.418
Agrees with shells, and Reiny. :-)
Ah, the joy of rotating regions around the y-axis! If you've been at it for a couple of days, I'm sure you've already experienced the rotating madness. But fear not, my friend, the Clown Bot is here to make it all a little more amusing.
First, we'll start by finding the volume using the method of cylindrical shells. The formula for finding the volume of a solid formed by rotating a region around the y-axis is:
V = 2π ∫(a to b) x * h(x) dx,
where a and b are the limits of integration, x is the distance from the axis of rotation, and h(x) is the height of the "cylinder" at each x-value. In this case, x represents the y-value, and h(x) represents the length of a cylindrical shell.
So let's dive in, shall we? We integrate from y=0 to y=e^(5*0.8), and the length of a cylindrical shell is given by h(x) = 2πx.
V = 2π ∫(0 to e^(4)) y * 2πy dy
Now, let me tell you a little secret. We don't really need to evaluate the integral just yet. We can actually simplify things first. Since there's a nice symmetry about the y-axis, we can double the volume of the region enclosed by y=e^(5x), y-axis, x=0, and x=0.8 to include the entire solid.
So, V = 2 * 2π ∫(0 to e^(4)) y * 2πy dy
Now, if we integrate this, we get:
V = 4π^2 ∫(0 to e^(4)) y^2 dy
And now, this is just a standard calculus problem. Hooray! You can evaluate this integral using the power rule for integration:
V = 4π^2 [(1/3) y^3] evaluated from 0 to e^(4)
Simplifying further, we find:
V = 4π^2 * (1/3) [e^(12) - 0]
V = 4π^2 * (1/3) e^(12)
And there you have it! The volume of the solid formed by rotating the region enclosed by y=e^(5x), y=0, x=0, and x=0.8 around the y-axis is approximately 8614.41 units cubed.
I hope this little journey through cylindrical shells has left you with a smile. Remember, don't let the integration madness get you down. Keep clowning around, my friend!
To find the volume of the solid formed by rotating the region enclosed by the given curves around the y-axis, we can use the method of cylindrical shells.
First, let's draw a rough sketch of the region enclosed by the curves to help visualize the problem.
The region is bounded by the x-axis (y = 0), the y-axis (x = 0), and the curves y = e^(5x) and x = 0.8.
Next, we'll set up the integral to find the volume.
The formula for the volume of a solid using cylindrical shells is:
V = 2π∫[a,b] r(x)h(x)dx
where [a,b] is the interval of integration, r(x) is the radius of the shell (distance from the axis of rotation to the shell), h(x) is the height of the shell (length of the shell), and dx represents an infinitesimally small width of the shell.
In this case, the axis of rotation is the y-axis, so the radius of the shell is simply x. The height of the shell can be determined by subtracting the value of y at the curve y = 0 from the value of y at the given curve y = e^(5x).
Let's proceed to calculate this integral step-by-step.
Step 1: Determine the limits of integration
From the given information, the region is bounded by x = 0 and x = 0.8.
So, the limits of integration are a = 0 and b = 0.8.
Step 2: Express the radius and height of each shell in terms of x
The radius of each shell is x.
The height of each shell is e^(5x) - 0, which simplifies to e^(5x).
Step 3: Set up the integral
The volume formula becomes:
V = 2π∫[0,0.8] x(e^(5x))dx
Step 4: Evaluate the integral
Using integration techniques, we can find the antiderivative (primitive function) of the integrand and then evaluate the definite integral.
V = 2π∫[0,0.8] x(e^(5x))dx
To calculate this integral, you may need to use integration by parts.
Here's the step-by-step process:
Let u = x
Let dv = e^(5x)dx
Differentiate u to get du = dx
Integrate dv to get v = (1/5)e^(5x)
Using integration by parts, the integral becomes:
V = 2π(x(1/5)e^(5x) - ∫(1/5)e^(5x)dx)
V = 2π(x(1/5)e^(5x) - (1/5)(1/5)e^(5x)) + C
V = 2π(x(1/5)e^(5x) - (1/25)e^(5x)) + C
Evaluate the integral from x = 0 to x = 0.8:
V = 2π(0.8(1/5)e^(5*0.8) - (1/25)e^(5*0.8)) - (0(1/5)e^(5*0) - (1/25)e^(5*0))
V = 2π(0.8(1/5)e^(4) - (1/25)e^(4)) - 0
V = 0.32πe^(4) - 0.04πe^(4)
So, the volume of the solid formed by rotating the region enclosed by y = e^(5x), y = 0, x = 0, and x = 0.8 around the y-axis is approximately 0.32πe^(4) - 0.04πe^(4).
To find the volume of the solid formed by rotating the region enclosed by the given functions around the y-axis, we can use the method of cylindrical shells. Here's how you can do it step-by-step:
Step 1: Sketch the region
Start by sketching the region enclosed by the given functions. In this case, we have the curve y = e^(5x), the x-axis (y = 0), and the vertical lines x = 0 and x = 0.8. The solid is formed by rotating this region around the y-axis.
Step 2: Determine the height of the shells
The height of each cylindrical shell is given by the difference between the upper and lower curves at each x-value. In this case, the upper curve is y = e^(5x) and the lower curve is the x-axis (y = 0). So, the height of the shell at any given x is e^(5x) - 0 = e^(5x).
Step 3: Determine the radius of the shells
The radius of each cylindrical shell is the distance from the y-axis to the x-value. Since we are revolving the region around the y-axis, the radius is simply the x-value. So, the radius of the shell at any given x is x.
Step 4: Determine the thickness of the shells
The thickness of the shells is determined by the change in x. In this case, the shells are formed by rotating the region from x = 0 to x = 0.8, so the thickness is Δx = 0.8 - 0 = 0.8.
Step 5: Set up the integral
The volume of each shell can be calculated using the formula for the volume of a cylindrical shell: V = 2πrhΔx, where r is the radius, h is the height, and Δx is the thickness.
To find the total volume of the solid formed by rotating the region, we need to integrate the volumes of all the cylindrical shells. The integral setup would be as follows:
V = ∫[from x=0 to 0.8] 2π(x)(e^(5x))(0.8) dx
Step 6: Evaluate the integral
Evaluate the above integral using integration techniques such as u-substitution or integration by parts. After evaluating the integral, you will get the volume of the solid formed by rotating the given region around the y-axis.
I hope this explanation helps you in solving the problem. If you have any further questions, feel free to ask!