It is always useful (and usually necessary) to graph the solid you are trying to find the volume of in order to determine the limits of integration. This can be done quite easily by hand in this case.
You can graph the plane by finding the intercept for each axis and then simply connecting those points.
Setting #x# and #y# equal to #0#, we find that the #z# intercept occurs at #z=3#. Similarly, we find that the #y# intercept occurs at #y=2# and the #x# intercept occurs at #x=4#.
The given solid is thus bounded below by the #xy#-plane (#z=0#), behind by the #yz#-plane (#x=0#), to the left by the #xz#-plane (#y=0#), and to the right by the plane #3x+6y+4z=12#.
Here is a 3-dimensional graph of the given plane:
Here is a graph of the complete solid:
In this case, you can choose to integrate with respect to any order of the variables. I have chosen to follow the order #dzdydx#. We can now determine the limits of integration.
If we were to enter the solid along the #z#-axis (or any path in the #z#-direction), moving from negative #z# to positive #z#, we would first hit the bottom of the solid, which is in the #xy#-plane, i.e. #z=0#. Thus, our lower limit of integration with respect to #z# is #0#. Continuing to travel along the #z#-axis, we would eventually exit the solid through that orange surface, which is the given plane. We can solve for #z# to determine the upper limit of integration.
#3x+6y+4z=12#
#z=1/4(12-3x-6y)#
Thus, our upper limit of integration with respect to #z# is #1/4(12-3x-6y)#, and the inner-most integral looks like this:
#int_0^(1/4(12-3x-6y))dz#
We have now finished with #z#, and so we can imagine pushing this solid down to the #xy#-plane to determine our remaining limits of integration, which forms a triangle (2-D). You can also continue to visualize the solid in 3-dimensions and simply set the #z# variable to #0# whenever it is present in an equation.
Now with respect to #y#, we would first enter the solid at #y=0# along the #y#-axis (or any path in the #y#-direction). You may visualize this as beginning at the origin in 2-dimensions or the #xz#-plane in 3-dimensions. Thus, our lower limit of integration with respect to #y# is #0#. Continuing to travel along a #y# path, we would exit the triangle through the line #3x+6y=12# (setting #z=0#). If you are visualizing in 3-dimensions, we would exit the solid through our plane, now with #z=0#. We can solve for #y# to find the upper limit of integration.
#3x+6y=12#
#y=1/6(12-3x)#
Thus, our upper limit with respect to #y# is #1/6(12-3x)#, and we now have a double integral of the form:
#int_0^(1/6(12-3x))int_0^(1/4(12-3x-6y))dzdy#
Finally, we can easily determine the upper and lower bounds for #x#. You can visualize this by pushing the triangle from the 2-dimensional orientation mentioned above to the #x#-axis. We can see that #x# will run from #0# to the point where the plane intersects the #x#-axis at #4#. Alternatively, you can set both #y# and #z# variables to #0# and solve the equation of the plane for #x#.
#3x=12#
#x=4#
Thus, our upper limit with respect to #x# is #4#, and we now have a triple integral of the form:
#int_0^4int_0^(1/6(12-3x))int_0^(1/4(12-3x-6y))dzdydx#
To find the volume of the solid, we keep the integrand at a value of #1#. We integrate with respect to #z# first, then #y#, then #x#.
#int_0^4int_0^(1/6(12-3x))int_0^(1/4(12-3x-6y))dzdydx#
#int_0^4int_0^(1/6(12-3x))zdydx#
Evaluating for our limits of integration from #0# to #1/4(12-3x-6y)#,
#int_0^4int_0^(1/6(12-3x))1/4(12-3x-6y)dydx#
We now integrate with respect to y:
#int_0^4(1/4(12y-3xy-(3y^2)))dx#
Evaluating for our limits of integration from #0# #1/6(12-3x)#,
#int_0^4(1/4[(12(1/6(12-3x))-3x(1/6(12-3x))-3(1/6(12-3x)^2)])dx#
#int_0^4(1/4((3x^2)/4-6x+12)dx#
#int_0^4(3x^2)/16-3/2x+3dx=4#
Thus, the volume of the solid is 4 #(units^3)#.
