Suppose that the region between the x-axis and the curve y=e^-x for x>=0 has been revolved around the x-axis. Find the surface area of the solid.
I got 3*pi
The book shows an answer of pi * [sqrt(2) + ln(1 + sqrt(2))]
Where do I go wrong? For the sides of the surface, I integrated 2*pi*e^-x over [0,inf] and got 2*pi. Then for the base of the surface, I pi*(e^0)^2 = 2*pi.
No, I don't see how you integrated that against x.
find a differential piece of the curve ds, so 2PI y ds is the surface area. Now change ds to dx, dy, where ds^2=dx^2+dy^2
Notice that dy=-y dx so
ds^2= dx^2 +Y^2 dx^2 = (1+y^2) dx^2
or ds=sqrt (1+y^2) dx
Now integrate with respect to x.
area= INT 2PI e^-x (1+e^-2x) dx from x=1 to inf
see if that helps.
Check my thinking, I haven't integrated it to check.
Thank you bobpursley.
My surface area integral was bad.
I was incorrectly assuming
S = Int 2*pi*f(x) dx
I read through proof. It is
S = Int 2*pi*f(x)*sqrt(1+(dy/dx)^2) dx
Thanks!
Gas is escaping from a spherical balloon at the rate of 2ft3/min. How fast is the
surface area shrinking when the radius is 2ft?
To find the surface area of the solid formed by revolving the region between the x-axis and the curve y = e^-x for x >= 0 around the x-axis, we can use the method of cylindrical shells.
First, let's clarify the approach you took. To find the surface area of the sides of the solid, you correctly integrated 2πe^-x over the interval [0, ∞]. However, to find the surface area of the base, you mistakenly used the formula for the area of a circle and substituted e^0 for the radius. This is incorrect since the base is not a circle, but a point at x = 0. Therefore, the base area should be zero.
Now, let's proceed with the correct method of finding the surface area using cylindrical shells.
The surface area of the solid obtained by revolving the region between the x-axis and the curve y = e^-x around the x-axis can be found using the formula:
Surface Area = ∫[a,b] 2πx * height(x) dx
Here, a and b represent the limits of integration, and the height(x) represents the distance from the curve to the axis of rotation at each point x.
In this case, the height(x) equals y = e^-x. Therefore, the surface area can be expressed as:
Surface Area = ∫[0,∞] 2πx * e^-x dx
To solve this integral, we can use integration by parts. Using the formula:
∫u * dv = uv - ∫v * du
Let's set u = x and dv = 2πe^-x dx. Differentiating u gives du = dx, and integrating dv gives v = -2πe^-x.
Now, applying the formula for integration by parts, we get:
Surface Area = [x * (-2πe^-x)] evaluated from 0 to ∞ - ∫[0,∞] (-2πe^-x) * dx
Evaluating the first part, we have:
[x * (-2πe^-x)] evaluated from 0 to ∞ = 0 - [0 * (-2πe^0)] = 0
Now, evaluating the second part, we have:
- ∫[0,∞] (-2πe^-x) * dx = 2π ∫[0,∞] e^-x dx
To solve this integral, we can use the fact that the integral of e^(-ax) is 1/a * e^(-ax), therefore:
2π ∫[0,∞] e^-x dx = 2π * [e^-x] evaluated from 0 to ∞
Applying the limits, we have:
2π * [e^-∞ - e^0]
Since e^-∞ equals zero, we have:
2π * [0 - e^0] = -2π
Finally, summing up the two parts, we have:
Surface Area = 0 + (-2π) = -2π
Therefore, the surface area of the solid is -2π, which does not match the answer you obtained or the one shown in the book. This suggests that either there may be an error in the question or the answer choices provided.