Thread: vol

taking slices at fixed y, how do i find the vol of the solid enclosed by y=x^2, y+z=2 and z=1?

You need a "floor" for your solid - I'm going to assume z=0.

You divide the solid into two parts - the first part, with y going from 0 to 1, has the z=1 plane on the top. The second part, with y going from 1 to 2, has the y+z=2 plane on the top. All of the cross sections are rectangles.





where I substituted length times width for the area in both cases. Of course, the total volume is .

Post again in this thread if you're still having trouble.

- Hollywood

may i know how ((2-y)-0) for V2 is obtained? i thought that the lenght for y would be (2-1)?

Also, how do you know that we can use a single integral to find the vol? i thought that double integrals are mean for areas and triple integrals are meant for vol?

thanks!

Originally Posted by alexandrabel90

may i know how ((2-y)-0) for V2 is obtained? i thought that the lenght for y would be (2-1)?

Also, how do you know that we can use a single integral to find the vol? i thought that double integrals are mean for areas and triple integrals are meant for vol?

thanks!

For , the "roof" is the plane z=1, but for , the "roof" is the plane y+z=2, which intersects the plane z=1 at y=1 and slopes down to intersect the "floor" z=0 at y=2. The height for any given y between 1 and 2 is given by z=2-y.

When I substituted length times width for the cross-sectional area, I actually did an (extremely easy) double integral. So that makes three integrals for a volume.

- Hollywood

does that mean that i can actually find volume of a surface via single, double or triple integrals?

Originally Posted by alexandrabel90

does that mean that i can actually find volume of a surface via single, double or triple integrals?

Yes. A single integral is a simplified double integral. A double intregal is a simplified triple integral. If you can see it right away, there is no need to use any more then 1 or 2.