Calculus 2 double

Question

Find the volume of the region above the plane and below the surface z=18 - 2x (square) - 2y (square)in three ways:

(a) By using a double integral in polar coordinates.

(b) By noticing that this is the volume of revolution of the region in the first quadrant

of the xz plane bounded by the curve z = 16 - x(square) and using disks method.

(c) By again viewing this as a volume of revolution, but using shells method

(d) If you are lucky,your work in part (a) should, somewhere along the way, contain the single integrals from either the disks or shells method. Circle the right step. If you are unlucky, you need to reverse the order of integration so that you do the 0 integral first before this works, so do that.

(e) deleted

(f) Use one of the methods above to calculate the work required to fill this region, pumping water from the ground. Assuming units are meters, so there are 1000km

of water per cubic meter, and the gravitational constant is g = 9.8. You only have to choose 1 method.

Answer 1

Part A:

We need to integrate f(x,y) = 18 - x^2 - y^2 over the circle x^2 + y^2 = 18, but using polar coordinates instead of cartesian coordinates. So we use the formula Int[a,b] ( Int[c,d] ( f(r,t) * r dr ) dt ), where c and d are the bounds for the radius r, and where a and b are the bounds for the angle t. Using the identity r^2 = x^2 + y^2, we can write f(r,t) = 18 - r^2, and since we are now in polar coordinates, we can see that the radius varies from r = 0 to r = sqrt(18) and the angle varies from t = 0 to t = 2*pi.

Thus the volume formula becomes:

V = Int[0, 2*pi] ( Int[0, sqrt(18)] ( (18 - r^2) * r dr ) dt )

= Int[0, 2*pi] ( Int[0, sqrt(18)] ( (18*r - r^3) dr ) dt )

= Int[0, 2*pi] ( 18*r^2/2 - r^4/4 evaluated from r=0 to r=sqrt(18) ) dt )

= Int[0, 2*pi] ( 81 ) dt )

= 81*t evaluated from t=0 to t=2*pi

= 162*pi.

Part B:

If we revolve the curve x = sqrt(18 - z) about the z-axis, then we can use the disc method to calculate this volume, where we are integrating the cross-sectional circular area with radius r = x(z) = sqrt(18-z), where z varies from z = 0 to z = 18.

Thus the volume formula becomes:

V = Int[0, 18] ( pi * (sqrt(18 - z)^2 dz )

= Int[0, 18] (pi * (18 - z) dz )

= pi * (18z - z^2/2 evaluated from z = 0 to z = 18 )

= 162*pi.

Part C:

If we instead partition the region into a series of concentric cylindrical shells with volume V = (surface area of shell) * (infinitesimal thickness) = (2*pi*x*f(x)) * (dx), then we can use the cylindrical shells method to calculate this volume, where we are integrating the surface area of each cylindrical shell given by the equation SA(x) = 2*pi*x*z(x) = 2*pi*x*(18-x^2), where x varies from x = 0 to x = sqrt(18).

Thus the volume formula becomes:

V = Int[0, sqrt(18)] ( 2*pi*x*(18 - x^2 dx )

= Int[0, sqrt(18)] ( 2*pi*(18x - x^3 dx )

= 2*pi*(18x^2/2 - x^4/4 evaluated from x = 0 to x = sqrt(18) )

= 2*pi*81

= 162*pi.

Part D:

V = 2*pi*Int[0, sqrt(18)] ( 18r - r^3 dr ).

V = pi*Int[0, 18] ( 18 - z dz ).

V = 2*pi*Int[0, sqrt(18)] ( 18x - x^3 dx ).

Part F:

Work is defined to be the integral of the force over a given distance, where the force in question is the force on each infinitesimal circular slice of water to be moved from ground into the volume. Since we have F = ma from newton's second law, we can use the density of water together with the volume of an infinitesimal circular slice of water to calculate the mass m. The value for the acceleration a is given by the problem to be g = 9.8 m/s^2.

Since the definition of density is p = m/v, then we have m = p*v, where p is the density of water given by p = 1000 kg/m^3. The volume of the circular slice is just pi*x(z)^2 * dz = pi*(sqrt(18 - z))^2 * dz = pi*(18 - z) * dz. The water must be moved from ground to a distance of z above the ground, and so the final work required to move an infinitesimal circular slice of water from z = 0 to some z between 0 and 18 is given by dW = mgz = (pv)gz = pg*pi * (18 - z)*z dz = 1000*9.8*pi * (18z - z^2) dz.

Thus the final work (in joules) integral becomes:

W = Int[0, 18] ( dW )

= Int[0, 18] ( 9800*pi*(18z - z^2) dz )

= 9800*pi*(18z^2/2 - z^3/3 evaluated from z=0 to z=18 )

= 9525600*pi Joules.