$\displaystyle \int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{3x^2+3z^2}^{45-2x^2-2z^2} ? dy dz dx$
I don't know what the equation is. Shouldn't it be $y_2 - y_1$?
$45-2x^{2}-2z^{2} - (x^{2}+3z^{2})$
? It's not being accepted, says: "Your answer isn't a number (it looks like a formula that returns a number)"
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Using a triple integral to find volume While it is beyond what is being asked here... if you had a density function ($\rho$) then mass = $V=\iiint \rho \,dz\,dy\,dx$ but this is just volume, so the density is equal to 1.
Now what about the examples that say $V = \iint y_2-y_1\, dz\,dx$?
They are the same thing!
The limits of y in the triple integral are $y_1$ to $y_2$. After the first integration.
$\iiint \,dy\,dz\,dx = \iint y_2-y_1\,dz\,dx$
As for the problem at hand. The contour where $y_2 - y_2 = 0$ defines the limits for $x$ and $z$.
$45 -3x^2-5z^2 = 0$
$x^2/15+z^2/9 = 1$
At which point I would be inclined to switch to quasi - cylindrical.
$x = \sqrt {15}\,r \cos \theta$, $z = 3r \sin \theta$
$dx\,dz = 3\sqrt{15}\, r\, dr \,d\theta$
$\int_0^{2\pi}\int_0^1 3\sqrt{15}\,(45-45r^2)r\,dr\,d\theta$
$(2\pi)(3\sqrt{15})(45)(1/2-1/4) $
If you are a masochist and you don't want to change coordinate system, (or you haven't learned how to do that yet.) $3x^2 + 5z^2 = 45$
$x = \pm \sqrt{15-5/3 z^2}$
$\int_{-3}^{3}\int_{-\sqrt{15-5/3 z^2}}^{\sqrt{15-5/3 z^2}} 45 - 3x^2 - 5z^2\, dx\,dz$
$\int_{-3}^{3} (45-5z^2)x - x^3 \,dz$
$\int_{-3}^{3} 2(45-5z^2)(15-5/3 z^2)^{1/2} - 2(15-5/3 z^2)^{3/2} \,dz$
$\int_{-3}^{3} 4(15-5/3 z^2)^{3/2}\,dz$
$z = 3\sin\theta, dz = 3\cos\theta \,d\theta$
$\int_{-\pi/2}^{\pi/2} 4(15-15\sin^2\theta)^{3/2}(3\cos\theta)\,d\theta$
$\int_{-\pi/2}^{\pi/2} 12(15)^{3/2}(cos^4\theta) \,d\theta$
$\int_{-\pi/2}^{\pi/2} 3(15)^{3/2}(1+cos\,2\theta)^2\,d\theta$
$\int_{-\pi/2}^{\pi/2} 3(15)^{3/2}(1+2cos\,2\theta + cos^2\,2\theta)\,d\theta$
$\int_{-\pi/2}^{\pi/2} 3(15)^{3/2}(1+2cos\,2\theta + 1/2(1+cos\,4\theta)\,d\theta$
$\int_{-\pi/2}^{\pi/2} 3(15)^{3/2}(3/2+2cos\,2\theta + (1/2)cos\,4\theta)\,d\theta$
When we integrate all of the $\cos n\theta$ terms become $(\sin n\theta)/n$ terms and they all equal 0.
$3(15)^{3/2}(3/2)(\pi)$
If you want the volume you would just compute $$\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{3x^2+3z^2}^{45-2x^2-2z^2} dydzdx$$ Notice how if you want the volume the function you're integrating just becomes one.
When you're doing a triple integral you are iterating over the entire 3-space of the solid so if you make the function one you are just then summing up solely the volume element $dV=dxdydz$ which in effect would give you the volume of the solid.
The situation you are thinking of is when you take a double integral. Say instead of doing a triple integral I was doing a double integral, I would want to integrate $$\int_{-3}^{3} \int_{-\sqrt {9-x^2}}^{\sqrt {9-x^2}} (y_2 - y_1)dzdx$$ like you were thinking. This is because $dxdz=dA$ gives a square, or an area, on the $xz $-plane, but we want volume. And so to get volume we just multiply that square base by its "height" in the y direction which is given by $y_2 - y_1$. But remember in a triple integral we are integrating over the entire 3-space, not just it's 2d projection. And notice how $dxdydz $ is a volume element, not an area element this time because it forms a cube in 3-space. So if our goal is still to get the volume we just have to sum up those small volume elements and nothing else, and that's why the function is one. (If your getting volume using a triple integral that function will always be 1, however if you want to get mass you may end up having a density function there.)
EDIT, RESPONSE TO COMMENTS
First for the trig substitution, we have $$\int_{-3}^{3} \frac{20\sqrt{(9-x^2)^3}}{3}dx$$ with $x=3\sin\theta$ and $dx=3\cos\theta\cdot d\theta$ we have $$\frac{20}{3}\int_{x=-3}^{x=3} 3\sqrt{(9-9\sin^2\theta)^3}\cos\theta\cdot d\theta=20 \int_{x=-3}^{x=3} \sqrt{9^3\cos^6\theta}\cos\theta\cdot d\theta$$ $$=20\cdot27\int_{x=-3}^{x=3}\cos^4\theta \cdot d\theta$$ which is easy to integrate. (To go from step one to two we just pull out the 9 and apply the trigonometric identity $\cos^2x=1-\sin^2x$).
Lastly on the volume, just remember that $dxdydz$ looks like an infinitely small cube in 3-space, so if we just sum up all these infinitesimal cubes within a region (which is what the integral does), we get the total volume of that region.