series((1+x)^n,x=0)

Sorry, the last two vertices should be (10,20) and (7,20). The Jacobian is (-3). Thus the integral becomes int(f(u,v)*abs(J),[u,v]=Rectangle(7..10,8..20)). Sorry for the mistakes in my 1st answer. Thank you, dr. Mihailovs, for noticing this! Steven.

Hello,

In my opinion, it is quite obvious that the solid common to two cilinders, having the same diameter, that cross each other at an angle of 90 degrees, is a sphere. How can you be sure that the solid we are looking for has the equation z^2 = 4 - max(x^2,y^2)?

Kind regards. Steven.

>with(plots):

> h := (x, y) -> piecewise(2 <= y and y <= 5, 4-x^2);

> plot3d(h(x, y), x = 0 .. 2, y = 0 .. 5, axes = boxed);

> with(VectorCalculus);

> int(h(x, y), x = 0 .. 2, y = 2 .. 5);

Kind regards. Steven.

First, I draw the intersection lines between the two cilinders:

f:=(x,y,z)->x^2+z^2-4:

g:=(x,y,z)->y^2+z^2-4;

with(plots):

> p1 := implicitplot3d([f(x, y, z), g(x, y, z)], x = -4 .. 4, y = -4 .. 4, z = -4 .. 4, axes = boxed);

> p2 := intersectplot(f(x, y, z), g(x, y, z), x = -4 .. 4, y = -4 .. 4, z = -4 .. 4, axes = boxed);

> display(p1, p2);

You see now that the solid common to both cilinders is a simple sphere with radius 2.

If you want a nice graph, then do this:

> p3:= implicitplot3d([f(x, y, z), g(x, y, z)], x = -4 .. 4, y = -4 .. 4, z = -4 .. 4, axes = boxed, style = wireframe, color = [blue, red]);

> p4 := implicitplot3d(x^2+y^2+z^2 = 4, x = -2 .. 2, y = -2 .. 2, z = -2 .. 2, color = yellow, style = surface);

> display(p3, p2, p4);

Kind regards. Steven.

First, I draw the intersection lines between the two cilinders:

f:=(x,y,z)->x^2+z^2-4:

g:=(x,y,z)->y^2+z^2-4;

with(plots):

> p1 := implicitplot3d([f(x, y, z), g(x, y, z)], x = -4 .. 4, y = -4 .. 4, z = -4 .. 4, axes = boxed);

> p2 := intersectplot(f(x, y, z), g(x, y, z), x = -4 .. 4, y = -4 .. 4, z = -4 .. 4, axes = boxed);

> display(p1, p2);

You see now that the solid common to both cilinders is a simple sphere with radius 2.

If you want a nice graph, then do this:

> p3:= implicitplot3d([f(x, y, z), g(x, y, z)], x = -4 .. 4, y = -4 .. 4, z = -4 .. 4, axes = boxed, style = wireframe, color = [blue, red]);

> p4 := implicitplot3d(x^2+y^2+z^2 = 4, x = -2 .. 2, y = -2 .. 2, z = -2 .. 2, color = yellow, style = surface);

> display(p3, p2, p4);

Kind regards. Steven.

Hello dooberman0320,

The integrand and the parallellogram are not randomly chosen. If you do the following substitution:

u = 2y-x

v=x+y

then the parallellogram has the vertices (7,8),(10,8),(10,10) and (7,10), f(u,v)=u^4/sqrt(v) and jacobian J = 1.

with(VectorCalculus):

int(f(u,v)*J,[u,v]=Rectangle(7..10,8..10));

I hope this will help you. Kind regards. Steven.

The limit of the given sequence is indeed infinity. The limit of the product n^(11/2)*g is the product of the limits. The limit lim(n^(11/2),n->infinity) goes to infinity, but the limit of the sum g does not go to zero, so it is impossible that the limit of the product goes to a finite number.

Kind regards.