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?

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- Apr 20th 2010, 02:54 PM #1

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- Apr 21st 2010, 06:07 AM #2

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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.

$\displaystyle V_1=\int_0^1A\ dy=\int_0^1(1-0)(\sqrt{y}-(-\sqrt{y})\ dy=\int_0^12\sqrt{y}\ dy$

$\displaystyle V_2=\int_1^2A\ dy=\int_1^2((2-y)-0)(\sqrt{y}-(-\sqrt{y})\ dy=\int_1^2(4-2y)\sqrt{y}\ dy$

where I substituted length times width for the area in both cases. Of course, the total volume is $\displaystyle V=V_1+V_2$.

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

- Hollywood

- Apr 21st 2010, 06:25 AM #3

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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!

- Apr 21st 2010, 07:41 AM #4

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For $\displaystyle V_1$, the "roof" is the plane z=1, but for $\displaystyle V_2$, 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

- Apr 21st 2010, 10:58 AM #5

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- Apr 21st 2010, 01:06 PM #6