Can't find the volume bounded between the two cylinders.

Question

I'm trying to find the volume bounded by: $$x^2+z=1$$ and $$y^2+z=1$$ in the first octant.

by projection in the $xy$ plane, my integral becomes: $\int _0^1\int _0^1\int _{1-y^2}^{1-x^2}\:dzdydx$ which evaluates to zero, which doesn't agree with my expectations.

Answer 1

By graphing the region in three-dimensions:

we can see that the region is bounded from above by both cylinders and elsewhere by the coordinate planes, we can also notice that the region is symmetric about $y=x$.

So we set up our triple integral as follows: $$\int _0^1\int _x^1\:\int _0^{1-y^2}\:dzdydx+\int _0^1\int _0^x\:\int _0^{1-x^2}\:dzdydx=2\int _0^1\int _0^x\:\int _0^{1-x^2}\:dzdydx$$ which evaluates to half in the end

Answer 2

Note that the volume under the surfaces $x^2+z=1$ and $y^2+z=1$ in the first octant is the same. Therefore, its integral can be evaluated as

$$V = 2\int_0^1dx\int_0^xdy\int_0^{1-y^2} dz= 2\int_0^1dx\int_0^x(1-y^2)dy = 2\int_0^1(x-\frac13 x^3)dx=\frac56$$

Answer 3

When graphed: this region is shown to have a symmetry about the like y = x. As a result, we can proceed by simply adding that as a region boundary, and multiplying the result by 2. This allows us to take the integral of one function of z as opposed to two. The region under a surface z bounded by x and y can be given by the general form $\ \int_{R}^{}\int_{R}^{}z dy dx $ where we take the y area of a function bounded by z and integrate that area with respect to the x area of a function bounded by z. Because we use symmetry, either of the two functions relating z would work, however they are not arbitrary as it will define the order of integration. Let's use the function $z=1-x^2$

We continue by setting up our bounds of integration:$\ \int_{R}^{}\int_{R}^{}1-x^2 dy dx $ it is crucial that the outer integral have numerical bounds in order to end up with an numerical integral so we proceed by first using the relation y = x for the integral with respect with y (our inner integral) saving our (0,1) for our outer integral x. This yields a final integral of:

2$\ \int_{0}^{1}\int_{0}^{x}(1-x^2) dy dx $

The computation is as follows

2$\ \int_{0}^{1}\int_{0}^{x}(1-x^2) dy dx $

2$\ \int_{0}^{1}\int_{0}^{x}(y-yx^2) |_{y=0}^{y=x} dx $

2$\ \int_{0}^{1}(x-\frac{x}{3}^3) dx $

2$\ ({\frac{x}{2}}^2-{\frac{x}{12}}^4) |_{0}^{1} $

2($\frac{1}{2} - \frac{1}{12}$)

$\frac{5}{6}$