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    I'm stuck on 1biii. I know the integral is V= (triple integral signs) dxdyz but I'm not sure how to work out the limits
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    (Original post by bobbricks)
    I'm stuck on 1biii. I know the integral is V= (triple integral signs) dxdyz but I'm not sure how to work out the limits
    I'd make the inner integral wrt z.

    You have one limit fixed, 1 or 2.

    The other comes from rearranging the equation of the plane into z=....

    Can you take it from there?
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    (Original post by bobbricks)
    I'm stuck on 1biii. I know the integral is V= (triple integral signs) dxdyz but I'm not sure how to work out the limits
    To add to the previous post:

    1. You should draw a diagram to get a feel for the volume to be traversed.

    2. Draw a picture of the shadow on the x-y plane of the region in question. Then choose suitable limits for x and y. One of these must have constants as upper and lower limits. You should imagine a thin strip parallel to either the x or y axes, and which is dragged from some constant value of x or y to another, thus covering the whole of the required x-y region. If you choose the strip parallel to the y-axis, then the coords of the top and bottom of the strip will be functions of x.

    3. Note that you want to "integrate out" a variable each time you evaluate one of the iterated integrals e.g. you want something like:

    \displaystyle V = \int_{x=a}^b \int_{y=f(x)}^{g(x)} \int_{z=f(x,y)}^c dz dy dx

    Here, when we integrate against z, we remove z, but leave a function of (x,y) for the y-integral; when we deal with the y-integral, we leave a function of x only for the x-integral, and the x-integral finally removes the x dependence as it has constant limits.
 
 
 
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