Double integral volume of a region bounded by two z planes

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To find the volume of the region bounded by the planes z = 1 - y^2 and z = y^2 - 1 for 0 ≤ x ≤ 2, one can split the volume into two integrals corresponding to each z function. It is important to determine the correct bounds for y, which should be independent of z. The discussion emphasizes sketching the surfaces to visualize the finite region of integration. The appropriate limits for y are suggested to be from (z + 1)^(1/2) to (1 - z)^(1/2), but clarity on the chosen region is crucial. Ultimately, the integration setup hinges on the relationship between the two z functions and the selected volume region.
mahrap
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a) find the volume of the region enclosed by
z = 1 - y^2 and z = y^2 -1 for x greater or equal to 0 and less than or equal to 2.

b) would i split up the volume into two integrals, each integral for each z function and then add them together? I also don't know how to find the bounds. Please help with thorough steps
 
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mahrap said:
b) would i split up the volume into two integrals, each integral for each z function and then add them together?
The simplest is to note that you can split it into two equal volumes (z > 0, z < 0), so you only have to calculate one. But in general, given z > f(x,y) and z < g(x,y), it's as easy to integrate g(x,y)-f(x,y).
Surfaces like this carve the space into several regions, some of which are infinite. You have to decide which (finite) region is being asked for. Best is to sketch it. In the present case, you have to choose between 1 - y2 < z < y2-1 and 1 - y2 > z > y2-1. Which do you think it is?
 
haruspex said:
The simplest is to note that you can split it into two equal volumes (z > 0, z < 0), so you only have to calculate one. But in general, given z > f(x,y) and z < g(x,y), it's as easy to integrate g(x,y)-f(x,y).
Surfaces like this carve the space into several regions, some of which are infinite. You have to decide which (finite) region is being asked for. Best is to sketch it. In the present case, you have to choose between 1 - y2 < z < y2-1 and 1 - y2 > z > y2-1. Which do you think it is?




When i sketch the surface it seems to me the limits of integration should be 0 to 2 for x and
(z+1)^(1/2) to (1-z)^(1/2) for y. would my function inside the integral be
( 1 - y^2 - (y^2 - 1) )
 
mahrap said:
When i sketch the surface it seems to me the limits of integration should be 0 to 2 for x and
(z+1)^(1/2) to (1-z)^(1/2) for y. would my function inside the integral be
( 1 - y^2 - (y^2 - 1) )
You need a range for y which does not involve z.
Pls answer my question: do you think you want the region 1 - y2 < z < y2-1 or 1 - y2 > z > y2-1? Once you've chosen that you'll see what the range for y is.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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