00:01
Hello students, to find the volume of the solid bounded between the plane z equal to 0 and z equal to 3, the cylinder y is equal to x square and y is equal to 2 minus x square, we can integrate the cross -sectional areas perpendicular to the z -axis over the z -range.
00:16
Let's set up integral in terms of z using the method of cylindrical cells.
00:20
The differential volume element is a cylindrical cell with thickness dz, radius r, height h.
00:26
The radius of the cell can be expressed in terms of y and the height of the cell is given by the difference between the upper and lower bounds of z.
00:34
The radius r of the cylindrical cell is given by the difference between the two curves that is y is equal to x square and y is equal to minus 2 minus x square.
00:43
Solving for x in terms of y for each curve, for y is equal to x square, x is equal to square root y and for y is equal to 2 minus x square, x is equal to y 2 minus y square root.
01:02
Squaring both to find the intersection point between the two curves, we set them equal to each other, that is square root y is equal to square root 2 minus y.
01:11
Squaring both sides of the equation will have y equal to 2 minus y, 2y equal to 2, so y equal to 1.
01:21
So the intersection points are, intersection points are 1, minus 1, 1, 1 1 and minus 1 1.
01:44
Now we can set up the integral that is b is from z 0 to z 3 and integration y equal to x square to 2 minus x square and from x equal to minus root y to root y dx dy dz.
02:06
Simplifying the integral will integrate with respect to x first, then y, then z.
02:10
So from x minus y to minus root over y to x equal to minus minus root y to root y 1 dx will have 2 root y.
02:24
So b is equal to, from z equal to 0 to 3, y equal to x square to 2 minus x square will have 2 root y dy dz...