find the volume when the 1st quadrant region bounded by y=sinx, y=cosx and x=0 is revolved around the x-axis and the y-axis.
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Consider a point (x,y) and a thin cylindrical shell of thickness dx which has the y-axis as its axis of rotation. The volume of the shell is the surface area times the thickness: 2(pi)xydx, where x is the radius of the cylinder, y its height, and dx the thickness of the shell. In this context, y=cos(x)-sin(x) and, if we change the limits to 0<x<(pi)/4 we get half the volume we need (because of two symmetrical halves), so, using S to denote integral and [low,high] to denote integration limits, the complete volume of rotation=4(pi)S[0,(pi)/4](x(cos(x)-sin(x))dx)=4(pi)S[0,(pi)/4]((xcos(x)-xsin(x))dx)=4(pi)S[0,(pi)/4](xcos(x)dx)-4(pi)S[0,(pi)/4](xsin(x)dx). The solid is symmetrical about the line x=(pi)/4 where the curves intersect one another where sin(x)=cos(x) (i.e., tan(x)=1).

Let u=x, dv=cos(x)dx, then du=dx and v=sin(x). d(uv)/dx=udv/dx+vdu/dx=xcos(x)+sin(x), and, integrating wrt x:

xsin(x)=S(xcos(x)dx)+S(sin(x)dx).

Thus, S(xcos(x)dx)=xsin(x)-S(sin(x)dx)=xsin(x)+cos(x).

Now, let dv=-sin(x)dx, then v=cos(x). So, uv=xcos(x)=S(-xsin(x)dx)+S(cos(x)dx).

Thus, S(-xsin(x)dx)=xcos(x)-sin(x).

We now have all we need to carry out the original definite integration:

Volume of rotation=4(pi)(xsin(x)+cos(x)+xcos(x)-sin(x))[0,(pi)/4]=

4(pi)((pi)/4sqrt(2)+(pi)/4sqrt(2)-1)=4(pi)((pi)/2sqrt(2)-1)=(pi)^2sqrt(2)-4(pi)=1.3914 cu units.

The volume of rotation about the x axis is found by considering thin discs of radius y, thickness dx, where y=cos(x)-sin(x). The volume of each thin disc is (pi)y^2dx. The solid is symmetrical about the line x=(pi)/4 as before. This time the integral is 2(pi)S[0,(pi)/4]((cos(x)-sin(x))^2)dx)=2(pi)S[0,(pi)/4]((1-2sin(x)cos(x))dx)=2(pi)S[0,(pi)/4]((1-sin(2x))dx)=2(pi)(x+cos(2x)/2)[0,(pi)/4]=2(pi)((pi)/4-1/2)=(pi)^2/2-1=3.9348 cu units.

by Top Rated User (651k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
83,025 questions
87,735 answers
1,969 comments
4,543 users