Find the volume of the solid bounded by the surfaces
Find the volume of the solid bounded by the surfaces: $$ z^2=xy,\qquad xy=a^2,\qquad xy=4a^2,\qquad x=2y,\qquad x=3y,\qquad z=0,\qquad(z>0). $$
Explanation of Method
The substitution $u=xy,\,v=\dfrac xy$ turns the four bounding curves into a coordinate rectangle $[a^2,4a^2]\times[2,3]$, and the height $z=\sqrt{xy}=\sqrt u$ becomes especially simple — making this a natural candidate for the change of variables, with Jacobian $|J|=\dfrac1{2v}$.
- Substitute $u=xy,\,v=x/y$ and find the Jacobian.
- Express $z$ in the new coordinates.
- Set up and evaluate the volume integral over the rectangle.
Step 1 · Substitution and Jacobian
With $u=xy,\,v=x/y$: $x=\sqrt{uv},\,y=\sqrt{u/v}$, giving Jacobian $\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|=\dfrac1{2v}$. The region becomes $u\in[a^2,4a^2],\,v\in[2,3]$.
Step 2 · Express $z$
$z=\sqrt{xy}=\sqrt u$.
Step 3 · Set up and evaluate
$$ V=\int_2^3\int_{a^2}^{4a^2}\sqrt u\cdot\frac1{2v}\,du\,dv = \frac12\left[\int_{a^2}^{4a^2}\sqrt u\,du\right]\left[\int_2^3\frac{dv}v\right]. $$
$$ \int_{a^2}^{4a^2}\sqrt u\,du = \frac23\big[(4a^2)^{3/2}-(a^2)^{3/2}\big] = \frac23(8a^3-a^3)=\frac{14a^3}3,\qquad \int_2^3\frac{dv}v=\ln\frac32. $$
$$ V = \frac12\cdot\frac{14a^3}3\cdot\ln\frac32 = \frac{7a^3}3\ln\frac32. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct evaluation of the $(u,v)$ double integral gives exactly $\dfrac{7a^3}3\ln\dfrac32$.
Result
The volume is: