[Math] Volume of solid under plane $4x-2y+z = 18$ and above the region bounded by $4x+y = 16$ and $x^{2}+y = 16$.

integrationmultivariable-calculus

Express the volume of the solid under the plane $4x-2y+z = 18$ and above the region bounded by $4x+y = 16$ and $x^{2}+y = 16$.

Can't determine lower y boundary!

Volume = $\displaystyle \int_0^4\int_0^{16-x^2} 18 -4x + 2y\,dy\,dx$

For some reason volume doesn't start being integrated from y = $0$.

Why?? It clearly should, judging by the quick sketch.

You basically need that red-shaded region (and then imagine there's a plane above that red-shaded region).

So y definitely starts from $0$… why not?

I mean, when x = $4$, y = $0$, no?

Best Answer

Here is a link to the graph of the region of integration.

Region bounded by graphs of $y=16-4x$ and $y=16-x^2$

Thus the lower limit should be $y=16-4x$ not $y=0$.

Related Solutions

Integration – Volume of Cube Section Above Intersection with Plane

For $1 \le \alpha \le 2$, it is much easier to visualize the integral by subtracting instead of adding pieces of the volume.

As shown in the picture below, when $1 \le \alpha \le 2$, the volume of cube section below intersection of the plane $x + y + z = \alpha$ is the difference of the volume of one big tetrahedron with width/height/depth $= \alpha$ with three smaller ones with width/height/depth $= \alpha-1$. So the volume above the intersection becomes

$$1 - \left( \frac16 \alpha^3 - 3 ( \frac16 (\alpha-1)^3 )\right) = 1 - \frac16 \alpha^3 + \frac12 (\alpha-1)^3$$

Cube section 1

Update

About the question whether this argument can be extended to higher dimension, the answer is yes. Let's look at the 3-dimension $2 \le \alpha \le 3$ case first.

Cube section 2

As one increases $\alpha$ beyond $2$, the three tetrahedron in first figure start overlap. As shown in second figure, the intersection of the three tetrahedra are now three even smaller tetrahedra of width/height/depth = $\alpha -2$.

Previous way to compute the "volume" of cube section below the plane $x + y + z = \alpha$ now subtract too much from this three even smaller tetrahedron. One need to add them back. As a result, the volume above the plane becomes:

$$\begin{align}&1 - \left( \frac16 \alpha^3 - 3(\frac16 (\alpha-1)^3 + 3(\frac16 (\alpha-2)^3 \right)\\ = & 1 - \frac16 \alpha^3 + \frac12 (\alpha-1)^3 - \frac12 (\alpha-2)^3\\ = & \frac{(3-\alpha)^3}{6} \end{align}$$

Let us switch to the $k$-dimension case. To compute the "volume" of the hypercube section above the hyperplane $x_1 + \ldots + x_k = \alpha$, the first step is to subtract the volume of a $k$-simplex of size $\alpha$ from 1.

if $\alpha \le 1$, we are done.
if $\alpha > 1$, we over subtract the volume of $\binom{k}{1}$ simplices of size $\alpha-1$ and need to add them back.
if $\alpha > 2$, the $\binom{k}{1}$ simplices of size $\alpha-1$ in step 2 intersect and the intersection is a union of $\binom{k}{2}$ $\;k$-simplices of size $\alpha-2$. This means in step 2, we have added back too much and need to subtract the volume again.

Repeat these arguments and notice in the middle of the process, we need to either add or subtract the volumes of $\binom{k}{i}$ $\;k$-simplices of size $\alpha-i$. The "volume" of interest finally becomes:

$$1 -\sum_{i=0}^{\lfloor \alpha \rfloor} (-1)^i \binom{k}{i} \frac{(\alpha-i)^k}{k!}$$

[Math] Compute the volume of a solid bounded by a cone and cylinder

The parameterized (polar) curve $$r(\theta) = 2 \sin \theta$$ can be written as the parameterized (Cartesian) curve $$(x(\theta), y(\theta)) = (2 \sin \theta \cos \theta, 2 \sin^2 \theta) = (\sin 2 \theta, \cos 2 \theta + 1);$$ since both of the components in the last expression have period $\pi$, the parameterization traces out the circular boundary twice over the interval $[0, 2 \pi]$; we only want to integrate over each direction $\theta$ once, so we should replace the interval over which we integrate $\theta$ by $[0, \pi]$.

We can see too why the triple integral you wrote evaluates to zero: The upper limit on the middle integral, is $2 \sin \theta$, and the symmetry of that function (together with the fact that $\theta$ appears nowhere else in the integral) ensures that the integral over $[\pi, 2\pi]$ has the same magnitude but the opposite sign as the desired integral over $[0, \pi]$.

Best Answer

Related Solutions

Integration – Volume of Cube Section Above Intersection with Plane

[Math] Compute the volume of a solid bounded by a cone and cylinder

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