How do I revolve the region bounded by y=xSinx and y=0 and x=pi and x=0 about the line x=5pi/4?

Jeremy - 2009-08-11 08:46:59 - Mathematics

I can find the what the graph will look like, but don't know where to start graphing a region that does not revolve around the x or y axis.

Best Answer:

I've drawn you a graph of this problem along with the first few lines of calculations, which I am repeating below. To see this graph, click on the following link: http://i369.photobucket.com/albums/oo133/gerryrains/CylindricalShellsAroundxequal5times.jpg Method of Cylindrical Shells Integrate 2*pi*r*h r = ((5*pi/4) - x) h = x*sin(x) π ∫ 2π * ((x * sin(x) * (5π/4) - x)) dx = 0 . . . . π 2π * ∫ ((5π/4)*x*sin(x)) - (x^2*sin(x)) dx . . . . 0 I have to get back to work now. Both x*sin(x) and (x^2)*sin(x) can be integrated using integration by parts. ∫ x*sin(x)dx Let u = x Then du = dx Let dv = sin(x)dx Then (integrating) v = -cos(x) So ∫ x*sin(x) dx = -x*cos(x) - ∫sin(x) dx = -x*cos(x) + cos(x)............Integral of x*sin(x) ---------------------------------- Now for x^2 * sin(x) Let u = x^2 Then du = 2x dx Let dv = sin(x) (Now Integrate) Then v = -cos(x) So ∫ x^2 * sin(x) dx = -(x^2 * cos(x)) - ∫ -2x*cos(x) dx Now we have to evaluate By now, I'm sure that you could finish the problem from here, but I'll do it :D Let u = -2x Then du = -2dx Let dv = cos(x) Then v (integrating) = sin(x) So ∫ -2x(cos(x)dx = -2x*sin(x) - ∫-2 * sin(x) = -2x*sin(x) - cos(x) So finally, ∫ x^2 * sin(x) dx = -(x^2 * cos(x)) - 2x*sin(x) - cos(x) <<--Integral of x^2 * sin(x) --------------------------------------------- Back to the original integral ((5π/4)*x*sin(x)) - (x^2*sin(x)) dx, we have (5π/4) * (-x*cos(x) + cos(x)) - -(x^2 * cos(x)) - 2x*sin(x) - cos(x) = (5π//4) * (-x*cos(x) + cos(x)) + (x^2)*cos(x) + 2x*sin(x) ------------------------------------------------------------------------- Now we have to evaluate this from 0 to π and then multiply the whole thing by (2π) At π this is [(5π//4) * (+π - 1)] - π^2 + 0 At 0 this is: [(5π/4) * (0-1)] + 0 + 0 = -(5π/4) This is (5 π^2 / 4) - (5π/4) + (5π//4) - π^2 After cancellation we get: (2π * ((5π^2 / 4) - π^2) = (5 * (π^3) / 2) - π^3) and since (5/2) - 1 = (3/2) we're finally done: (3π^3 / 2)....<<<<<.....Answer .

Answer:

I've drawn you a graph of this problem along with the first few lines of calculations, which I am repeating below. To see this graph, click on the following link: http://i369.photobucket.com/albums/oo133/gerryrains/CylindricalShellsAroundxequal5times.jpg Method of Cylindrical Shells Integrate 2*pi*r*h r = ((5*pi/4) - x) h = x*sin(x) π ∫ 2π * ((x * sin(x) * (5π/4) - x)) dx = 0 . . . . π 2π * ∫ ((5π/4)*x*sin(x)) - (x^2*sin(x)) dx . . . . 0 I have to get back to work now. Both x*sin(x) and (x^2)*sin(x) can be integrated using integration by parts. ∫ x*sin(x)dx Let u = x Then du = dx Let dv = sin(x)dx Then (integrating) v = -cos(x) So ∫ x*sin(x) dx = -x*cos(x) - ∫sin(x) dx = -x*cos(x) + cos(x)............Integral of x*sin(x) ---------------------------------- Now for x^2 * sin(x) Let u = x^2 Then du = 2x dx Let dv = sin(x) (Now Integrate) Then v = -cos(x) So ∫ x^2 * sin(x) dx = -(x^2 * cos(x)) - ∫ -2x*cos(x) dx Now we have to evaluate By now, I'm sure that you could finish the problem from here, but I'll do it :D Let u = -2x Then du = -2dx Let dv = cos(x) Then v (integrating) = sin(x) So ∫ -2x(cos(x)dx = -2x*sin(x) - ∫-2 * sin(x) = -2x*sin(x) - cos(x) So finally, ∫ x^2 * sin(x) dx = -(x^2 * cos(x)) - 2x*sin(x) - cos(x) <<--Integral of x^2 * sin(x) --------------------------------------------- Back to the original integral ((5π/4)*x*sin(x)) - (x^2*sin(x)) dx, we have (5π/4) * (-x*cos(x) + cos(x)) - -(x^2 * cos(x)) - 2x*sin(x) - cos(x) = (5π//4) * (-x*cos(x) + cos(x)) + (x^2)*cos(x) + 2x*sin(x) ------------------------------------------------------------------------- Now we have to evaluate this from 0 to π and then multiply the whole thing by (2π) At π this is [(5π//4) * (+π - 1)] - π^2 + 0 At 0 this is: [(5π/4) * (0-1)] + 0 + 0 = -(5π/4) This is (5 π^2 / 4) - (5π/4) + (5π//4) - π^2 After cancellation we get: (2π * ((5π^2 / 4) - π^2) = (5 * (π^3) / 2) - π^3) and since (5/2) - 1 = (3/2) we're finally done: (3π^3 / 2)....<<<<<.....Answer .