Let $E$ be the solid region bounded by two surfaces, $z=10-r^2$ and $z=2+r^2$ in cylindrical coordinates. I computed the volume a lot of times using cylindrical coordinates and got $\frac{64\pi}{3}$. But the answer said $16\pi$. Which one is correct?
Volume of a solid in cylindrical coordinates
calculusintegrationmultivariable-calculus
Related Solutions
Answer using Cylindrical Coordinates:
Volume of the Shared region =
Equating both the equations for z, you get z = 1/2. Now substitute z = 1/2 in in one of the equations and you get r = $\sqrt{\frac{3}{4}}$.
Now the sphere is shifted by 1 in the z-direction, Hence
Volume of the Shared region = $$\int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{3}{4}}} \int_{1-\sqrt{1-r^2}}^{\sqrt{1-r^2}} rdzdrd\theta$$
$$V=2\pi \int_{0}^{\sqrt{\frac{3}{4}}} [2{\sqrt{1-r^2}}-1] rdr$$
substitute $$u = 1-r^2 ; r = 0 => u = \frac{1}{4} ; r = \sqrt{\frac{3}{4}} => u = 1$$
$$V = 2\pi [-\int_1^{\frac{1}{4}} u^{\frac{1}{2}} du - \int_{0}^{\sqrt{\frac{3}{4}}} rdr]$$
$$V= 2\pi (\frac{2}{3}u^{\frac{3}{2}}) - (\frac{r^2}{2})$$ $$V =2\pi*( \frac{2}{3}(1-\frac{1}{8}) - \frac{3}{8})$$
$$V = 2\pi*(\frac{14}{24} - \frac{3}{8}) = 2\pi*\frac{5}{24} = \frac{5}{12} \pi$$
I will only give an answer for the Cartesian method, to help you on your way.
For both Cartesian and Cylindrical Polar methods, you need to integrate between $z = 0$ to $z = 6$ the cross section $x^2 + y^2 = 1, y≥\frac{1}{2}$
In Cartesian, we clearly have $\frac{1}{2} ≤ y ≤1$ and then $-\sqrt{1-y^2}≤x≤\sqrt{1-y^2}$ from our equation of the circle. So our integral is thus: $$\int_{z=0}^6\int_{y=\frac{1}{2}}^1\int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}}\mathrm{d}x\mathrm{d}y\mathrm{d}z$$ I recommend making the substitution $y=\sin\theta$ to solve the integrand with the square root.
The Cylindrical Polar method is very similar: just remember your Jacobian.
Best Answer
Use back usual cartesian coordinates (to understand, perhaps, better your solid), so you surfaces are the ("going down") paraboloid $\;z=10-x^2-y^2\;$ and the ("going up") paraboloid $\;z=2+x^2+y^2\;$.
These two "bowls" meet at the circle $\;10-x^2-y^2=2+x^2+y^2\implies x^2+y^2=4\;$, so projecting onto the plane $\;z=0\;$, we get that the volume is
$$V=\int_{-2}^2dx\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\left((10-x^2-y^2-(2+x^2+y^2)\right)$$
If the above is what you did, in whatever coordinates, then you shall get the correct answer. Now, the above in cylindrical coordinates is
$$\int_0^2dr\int_0^{2\pi}\overbrace{r}^{\text{jacobian}}(10-r^2-(2+r^2))d\theta=\left.2\pi\int_0^2r(8-2r^2)dr=-\frac\pi4(8-2r^2)^2\right|_0^2=16\pi$$