Question:
I'll get an ice cream cone as a gift of answering the tricky question from mt husband. My husband asks me to evaluate the volume of the gift before having it. Assume the ice cream is inside the equation $2x^2+2y^2+2z^2=9$ and above the cone $3z^2=2x^2+2y^2$
My attempt,
$z^2=\frac{9}{2}-x^2-y^2$ is my first equation
$3z^2=2x^2+2y^2$ is my second equation.
So I found $x^2+y^2=\frac{27}{10}$ which made my $z=\frac{3}{\sqrt{5}}$
Volume$=\pi \int_{\frac{3}{\sqrt 5}}^{\frac{3}{\sqrt2}} x^2 dz$
$=\pi \int_{\frac{3}{\sqrt 5}}^{\frac{3}{\sqrt2}} \frac{9}{2}-z^2 dz$
$=\pi(\frac{9\sqrt{2}}{2}-\frac{117\sqrt{5}}{50})$
Is my answer correct? Is there another way to solve it?
No, it's not correct, and quite frankly I can't even understand what you're doing there.
The solid region $R$ that you have to integrate upon is the intersection of a ball (of radius $\dfrac 3 {\sqrt 2}$) and the upper part of a solid cone (in the following, I shall consider the $x$ axis to point toward you, the $y$ axis to the right, and $z$ axis upwards). You have to compute $\int _R 1 \ \Bbb d x \Bbb d y \Bbb d z$.
Since both the cone and the ball have the $z$ axis as axis of symmetry, the standard procedure is to pass to cylindrical coordinates:
$$\left\{ \begin{eqnarray} x &=& r \cos t \\ y &=& r \sin t \\ z &=& z \end{eqnarray} \right. , \ r \in (0,\infty), \ t \in (0, 2\pi) .$$
The equations of $R$ become
$$\left\{ \begin{eqnarray} 2r^2 + 2z^2 &\le& 9 \\ 2r^2 &\le& 3z^2 . \end{eqnarray} \right.$$
It is essential for the following that we find the height at which the cone intersects the sphere. To do this, why solve for $z$ the system $$\left\{ \begin{eqnarray} 2r^2 + 2z^2 &=& 9 \\ 2r^2 &=& 3z^2 . \end{eqnarray} \right.$$
This gives the equation $5z^2 = 9$ with the solution $z = \dfrac 3 {\sqrt 5}$ (we discard the other solution because it is negative). The $r$ corresponding to this $z$ is given by $r = \sqrt {\dfrac {27} {10}}$ (the other solution is negative).
Notice that $R$ is made up of two solid subregions. One is $R_1$, the piece of cone from height $z=0$ to height $z = \dfrac 3 {\sqrt 5}$. The other one is $R_2$, the piece of ball (like a round dome) that sits atop the plane $z = \dfrac 3 {\sqrt 5}$. Consequently, we shall have $\int _R 1 = \int _{R_1} 1 + \int _{R_2} 1$.
The first integral is just the volume of the cone of height $\dfrac 3 {\sqrt 5}$ and radius $\sqrt {\dfrac {27} {10}}$. You either know the formula, or you compute it with
$$\int \limits _{R_1} 1 \ \Bbb d x \Bbb d y \Bbb d z = \int \limits _0 ^{2\pi} \left[ \int \limits _0 ^{\frac 3 {\sqrt 5}} \left( \int \limits _0 ^{\frac 3 2 z} r \ \Bbb d r \right) \Bbb d z \right] \Bbb d t .$$
The upper bound in the innermost integral come from the inequation $2r^2 \le 3z^2$ corresponding to the solid cone.
For the spherical dome,
$$\int \limits _{R_2} 1 \ \Bbb d x \Bbb d y \Bbb d z = \int \limits _0 ^{2\pi} \left[ \int \limits _{\frac 3 {\sqrt 5}} ^{\frac 3 {\sqrt 2}} \left( \int \limits _0 ^{\sqrt {\frac 9 2 - z^2}} r \ \Bbb d r \right) \Bbb d z \right] \Bbb d t . $$
The upper bounds in the innermost integral come from the inequation $2r^2 + 2z^2 \le 9$ corresponding to the ball.
In both integrals, the outermost integral is trivially $2\pi$. Don't forget to add up the two results.