# Calculate the volume of the solid bounded by the region

Using spherical co-ordinates, find the volume of the solid region bounded above by the cone $$z^2=3(x^2+y^2), z\geq 0$$ and bounded below by the sphere $$x^2+y^2+z^2=4.$$

I tried in the following way: $$\displaystyle \int_0^{2\pi} \int_0^{\pi/6} \int_{0}^{2} \rho^2 \sin\phi \ d\rho \ d\phi \ d\theta$$

$$\displaystyle = \int_0^{2\pi} \int_0^{\pi/6} \left[\cfrac{\rho^3}{3}\right]_0^2 \sin\phi \ d\phi \ d\theta$$

$$\displaystyle = \cfrac{8}{3} \int_0^{2\pi} \left[- \cos \phi\right]_0^{\pi/4} \ d\theta$$

$$\displaystyle = \cfrac{16\pi}{3} \left(1 - \cfrac{\sqrt 3}{2} \right)$$

Is my solution is correct?

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Nov 16, 2023 at 7:24
• I am getting problems in finding limits of the integration. Nov 16, 2023 at 7:33
• Your comment does not add anything to your post. Please again: avoid "no clue" questions: "include your work and thoughts on the problem". Nov 16, 2023 at 9:15
• I have added my solution.Just wanted to know this is correct. Nov 16, 2023 at 10:40

Your solution is not correct because the interval for $$\phi$$ is not $$[0,\pi/6]$$ but $$[\pi/6,\pi].$$ The answer is therefore: $$V=\frac{16\pi}3\left[-\cos\phi\right]_{\pi/6}^\pi=\frac{16\pi}3\left(1+\frac{\sqrt3}2\right).$$ Note that the cone bounds your solid above, but the sphere does not bound it "below": rather "around". If you meant the other way round, the answer is of course $$\frac43\pi2^3-V=\frac{32\pi}3-V=\frac{16\pi}3\left(1-\frac{\sqrt3}2\right).$$
I don't understand why you said 'get the limits of the integration...', the volume is finite. $$I=\iint_{D}\left(\sqrt{4-x^2-y^2}-\sqrt{3(x^2+y^2)}\right)\mathrm{d}x\mathrm{d}y$$ Here $$D$$ is the $$xy$$-region determined the intersection. It should be easy, I left it to you.
When $$\sqrt{4-r^2}=\sqrt{3r^2}$$, the upper bound of $$r$$ is $$1$$. So$$I=\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{1}(\sqrt{4-r^2}-\sqrt{3r^2})r\mathrm{d}r=\frac{16\pi}{3}-\frac{8\sqrt{3}\pi}{3}$$
EDITED: I misunderstood your meaning, the result should be volume of ball substracts $$I$$. $$\frac{32\pi}{3}-I$$
• $I=\frac{16\pi}{3}-\frac{8\pi}{\sqrt{3}}$ Nov 16, 2023 at 10:19
• 1) The OP wanted to use spherical coordinates, which are more convenient for his problem. 2) Your $I$ does not at all represent the volume of his solid, which is bounded above by the cone and "below" by the sphere. Nov 16, 2023 at 11:33