Let R be the region in the first quadrant enclosed by the lines x= ln 3 and y = 1 and the graph of y= e^x/2.?
Assuming you mean y = e^(x/2): Since y = -1 is parallel to the x-axis, if we keep the functions in terms of x, we should use the disk/washer method. Take a vertical strip at some x in (0, 1) between y = 1 and y = e^(x/2) [draw it!]. Rotating this about the line y = -1 generates a washer with: (i) Bigger radius with length e^(x/2) - (-1) [via 'top - bottom'], and (ii) Smaller radius with length 1 - (-1). Since the area of a circle is πr^2, we obtain: V = ∫(x = 0 to ln 3) [π(e^(x/2) - (-1))^2 - π(1 - (-1))^2] dx. Evaluating this yields ∫(x = 0 to ln 3) π [(e^(x/2) + 1)^2 - 2^2] dx = π ∫(x = 0 to ln 3) [(e^x + 2e^(x/2) + 1) - 4] dx = π ∫(x = 0 to ln 3) (e^x + 2e^(x/2) - 3) dx = π (e^x + 4e^(x/2) - 3x) {for x = 0 to ln 3} = π [(3 + 4 * 3^(1/2) - 3 ln 3) - (1 + 4)] = π (4√3 - 3 ln 3 - 2). I hope this helps!
Find volume of solid generated by revolving region in first quadrant bounded by the coordinates axes,?
We can divide the region into vertical or horizontal strips. Using vertical strips of length y and width dx, if such a strip of area ydx at x is rotated about the line x=ln2 then it forms a cylindrical shell of radius r=ln2-x and volume dV = 2πr.ydx = 2π(ln2-x)e^xdx. Suppose the integral has the form V(x)=2π(a+bx)e^x. Then we need dV/dx = 2π[be^x + (a+bx)e^x] = 2π((a+b)+bx)e^x. This should equal 2π(ln2-x)e^x. Equating like terms : b = -1 a+b = ln2 a = ln2+1. Therefore V(x) = 2π(ln2+1-x)e^x. The limits of integration are from x=0 to x=ln2 : V = 2π[(ln2+1-ln2)e^ln2 - (ln2+1-0)e^0] = 2π(1 - ln2) = 1.928 cu units.
Let R be the region in the first quadrant enclosed by the graphs of f(x)=8x^3 and g(x)=sin(pi x), as shown...?
(a) Differentiate f(x) and plug in 1/2 for x to get the slope of your line. Plug 1/2 into f(x) to find a point on the line. Using point slope form, the slope and a point is sufficient to write an equation for the line. (b) Set f(x) equal to g(x) and solve for x. You should get two values. Using those values as the limits of integration find the integral of g(x)-f(x). (c) Using the same limits of integration as in (b) find the integral of (pi)(1-f(x))^2-(pi)(1-g(x))^2 Note if either (b) or (c) comes out negative take the absolute value of your answer.
What is the volume generated by rotating the area bounded by y=e^(2x), the y axis and the line y=2 about the x axis?
Using the disc method:First we find the intersection:[math]2=e^{2x}[/math][math]\ln 2 = 2x[/math][math]x = \dfrac{\ln 2}{2}[/math]Now we integrate:[math]\displaystyle V=\pi \int_0^{\frac{\ln 2}{2}} 2^2 - \left(e^{2x}\right)^2 dx[/math][math]\displaystyle V=\pi \ln 4 - \pi \int_0^{\frac{\ln 2}{2}} e^{4x} dx[/math][math]\displaystyle V=\pi \ln 4 - \pi (1-\frac{1}{4})[/math][math]\displaystyle V=\pi(\ln 4 - \frac{3}{4}) [/math]I purposely did the integration fast because you surely need more practice in it, and if you turned in your assignment as I did it, you would probably not get full marks. If you can't figure out how I did the integration, leave a comment and I can give some hints.
What is the volume of the solid generated by revolving the region bounded x=siny, x=1, and x-axis about the y-axis?
I will assume that the questing meant for the upper and lower bounds to be y=1 and the x-axis, as x = 1 wouldn’t provide a very well defined solid (it could be infinite, 0, have gaps, etc.). However, if you did intend to say x=1, I will assume you want the upper limit of the integral to be the first time x=1 and [math]x=\sin(y)[/math] intersect, at [math]y= \frac{\pi}{2}[/math]. If so, just replace the upper limit of 1 with [math]\frac{\pi}{2}[/math] in the final steps.Whenever we have to revolve an area around the x-axis to create a 3D solid, we simply make a composite function, [math]f(y)=v(g(y))[/math], where [math]v(r) = \pi r^2[/math] and in this case [math]g(y) = \sin(y)[/math][math]f(y) = \pi \sin^2y[/math]Now we just integrate this.[math]\displaystyle \int \pi \sin^2y dy[/math][math]\pi \int \sin^2y dy[/math]because [math]sin^2\theta = \frac{1 - \cos(2\theta)}{2}[/math][math]\pi \int \frac{1-\cos(2y)}{2} dy[/math][math]\frac{\pi}{2} \int 1-\cos(2y) dy[/math][math]\frac{\pi}{2}(\int 1dy - \int \cos(2y) dy)[/math][math]\frac{\pi}{2}(y - \int \cos(2y) dy)[/math]Let [math]u = 2y[/math]Then [math]\frac{du}{dy} = 2[/math] and [math]dy = \frac{du}{2}[/math]Substitute.[math]\frac{\pi}{2}(y - \int \frac{\cos(u)}{2} du)[/math][math]\frac{\pi}{2}(y - \frac{1}{2}\int \cos(u) du)[/math][math]\frac{\pi}{2}(y - \frac{1}{2}\sin(u))[/math]And substitute 2y back in.[math]\frac{\pi}{2}(y - \frac{\sin(2y)}{2})[/math]Now we can evaluate this at our upper and lower limits.[math]\frac{\pi}{2}((1) - \frac{\sin(2(1))}{2}) - \frac{\pi}{2}((0) - \frac{\sin(2(0))}{2})[/math][math]\frac{\pi}{2}(1 - \frac{0.909}{2}) - 0[/math][math]\boxed{= 0.857}[/math]
Find the volume of the solid obtained by rotating the region bounded by the given curves?
Shell method: V = 2π ∫ r h dr {a,b} .......... limits in {} radius is (1+y) to rotate around y=-1 = 2π ∫(1+y)(2) dx {0,1/3} + 2π ∫(1+y)(1/y - 1) dx {1/3,1} = 4π [y + y²/2] {0,1/3} + 2π [ ln(y) - y²/2 ] {1/3,1} = 14π/9 + 2π [ ln(3) - 4/9 ] = 2π [ ln(3) + 1/3 ] ≈ 8.997 units³ Disc method: V = π ∫ f(x)² dx {a,b} rotate around y=-1 extends the inner and outer radii by 1 = π ∫ (1/x + 1)² - (1)² dx {1,3} .............. note: π [(Ro)² - (Ri)²] = π ∫ 1/x² + 2/x dx {1,3} = π [ -1/x + 2 ln(x) ] {1,3} = π [ -1/3 + 2 ln(3) + 1 ] = 2π [ ln(3) + 1/3 ] Answer: ≈ 8.997 units³
What is the Volume of arctanx rotated about the y-axis?
The question makes little sense as stated. The arctan function is not a geometrical object, it is a mapping. Its graph, however, can be treated as a geometrical object, but rotating it gives a surface, so its volume is zero.I'm going to make an outlandish and unjustifiable suggestion, that you might possible be wondering about the volume created by rotating the curve about the y-axis.On the other hand, perhaps you mean the area of the surface. I'll do the volume first.The arctan function is multivalued. Let's restrict it to the branch where y is between 0 and pi/2. (Negative values of y are uninteresting because the curve is identical to the positive part apart from sign). As the curve goes out to infinity, it is initially possible that the area is infinite, so until proved otherwise I will consider the area when 0 < y < A < p/2. Later we can see what happens as A tends to pi/2 from below.The volume is the same as that obtained by rotating the area under the tangent curve, y = tan x, about the x axis and this might help to put the question on more familiar ground. Divide the volume between x = 0 and x = A into discs perpendicular to the x axis. The required volume is the integral of pi y^2 dx = pi tan^2 x dx from x = 0 to A.Because tan^2 x = sec^2 x - 1, its indefinite integral is tan x - x and the volume is pi(tan A - A). As A tends to pi/2 the volume becomes infinite as we first feared.For the area of the surface between the same limits, we need to integrate the rotated surface elements, i.e we need to integrate 2pi ds = 2pi sqrt(1+y'^2) dx. This simplifies to 2pi sec x dx with integral 2pi ln(sec x + tan x) so the area is 2pi ln(sec A + tan A) which tends to infinity as A tends to pi/2.Your ancestor, Joseph, would probably have done this in his head, whereas maybe it's doing your head in.