Complex value for volume, using triple integrals

Question

I'm trying to calculate the volume a hyperboloid, within $$z=0$$ and $$z+\frac 12 x-3=0.$$

The hyperboloid: $$x^2+\left(\frac y2\right)^2-z^2=5.$$

I calculated the projections on $xz$, $yz$, to use as bounds. $$\int_{-2\,\sqrt {5}}^{2\,\sqrt {5}}\!\int_{-\sqrt {5}}^{\sqrt {5}}\! \int_{1/2\,\sqrt {4\,{x}^{2}+{y}^{2}-20}}^{3-x/2}\!1\,{\rm d}z \,{\rm d}x\,{\rm d}y=-{\frac {40\,\sqrt {5}}{3}}+{\frac {40\,\sqrt {5} \ln \left( 2 \right) }{3}}+120-{\frac {20\,i}{3}}\sqrt {5}\pi $$

This result is complex, so i'm guessing i'm wrong.. can someone help me out with this one?

Maple:

restart;
with(plots):


hyp:=x^2+(y/2)^2-z^2=5:
vlak1:=z+1/2*x+3=0:
vlak2:=z+1/2*x-3=0:

f(x,y):=solve(vlak1,z):
g(x,y):=solve(vlak2,z):
h(x,y):=solve(hyp,z)[1]:
yzProjectie:=subs(x=0,h(x,y)):
yzSlope:=subs(x=0,g(x,y)):
xzSlope:=subs(y=0,g(x,y)):

xzProjectie:=subs(y=0,h(x,y)):
yzProjectie:=subs(x=0,h(x,y)):

yBounds:=solve(yzProjectie=0):
xBounds:=solve(xzProjectie=0):
Int(Int(Int(1,z=h(x,y)..g(x,y)),x=xBounds[2]..xBounds[1]),y=yBounds[2]..yBounds[1]);

Answer 1

In $(x,y,z)$-description the body $B$ in question is more complicated than your setup of nested integrals is suggesting. In particular, $B$ contains points $(x,y,z)$ with $|x|>\sqrt{5}$.

In order to simplify matters I propose the following:

When we intersect the (infinite) hyperboloid with parallel planes $z=c-{x\over2}$ we obtain ellipses $E_c$ with one main axis in the $(x,z)$-plane and the other main axis parallel to the $y$-axis.

From $$x^2-5=z^2=\left(c-{x\over2}\right)^2$$ we obtain the $x$-coordinates of the endpoints of one of these axes: $$x={2\over3}\bigl(-c\pm\sqrt{15+4c^2}\bigr)\ .$$ It follows that the corresponding semiaxis has length $$a={2\over3\cos\phi}\sqrt{15+4c^2}\ ,$$ whereby $\phi$ denotes the tilting angle of our planes with respect to the horizontal. The center of the ellipse $E_c$ has coordinates $\left(-{2c\over3},0,{4c\over3}\right)$, so that the endpoints of the other main axis satisfy $${4c^2\over9}+{y^2\over4}-{16c^2\over9}=5\ .$$ Therefore the corresponding semiaxis has length $$b={2\over\sqrt{3}}\sqrt{15+4c^2}\ ,$$ and the area of the ellipse $E_c$ is given by $$|E_c|=\pi a b={4\pi\over3\sqrt{3}\cos\phi}(15+4c^2)\ .$$ Between two ellipses $E_c$ and $E_{c'}$ there is a plate of thickness $|c-c'|\cos\phi$. We therefore can say that $$2{\rm vol}(B)=\int_{-3}^3 |E_c|\>\cos\phi\>dc={4\pi\over3\sqrt{3}}\int_{-3}^3(15+4c^2)\>dc$$ (here a symmetry argument came in), so that we finally obtain $${\rm vol}(B)=36\sqrt{3}\>\pi\ .$$

Answer 2

Your limits of integration in $x$ and $y$ cover a rectangle, while the projection of your solid to the $(x, y)$-plane is the ellipse with equation $$ x^{2} + \tfrac{1}{4}y^{2} - (3 - \tfrac{1}{2}x)^{2} = 5. $$ (That isn't totally obvious, incidentally. You need to inspect how the plane $z = \frac{1}{2}x - 3$ intersects the hyperboloid versus how the plane $z = 0$ intersects the hyperboloid.)

If you haven't done many examples of this type before, you'll likely find the computation easier to follow if you first introduce $v = \frac{1}{2}y$, work in cylindrical coordinates in $(x, v, z)$-space, and halve the resulting volume (to correct for the volume doubling of the change of coordinates).