0
$\begingroup$

Find the volume of the region B bounded above by the sphere $x^2 + y^2 + z^2 = a^2$ and below by the plane $z = b$, where $a > b > 0$.

Here I worked out $z=\sqrt{a^2-r^2}$ and $x^2+y^2=a^2-b^2$ so $0≤r ≤\sqrt{a^2-b^2}$ and $0≤θ≤2π$ and $\int_{0}^{2\pi }\int_{0}^{\sqrt{a^2-b^2}} r(a^2-r^2)^\frac{1}{2}drd\theta$ = $(0.5(a^2-b^2)a^2-0.25(a^2-b^2)^2)2π$ but the book has a different answer. I dont know where I have gone wrong i havent made any mistakes in the calculation.

Edit: I think they have used cylindrical coordinates with triple integral whereas I've used polar with doouble integral but i thought it doesnt make a difference?

$\endgroup$
  • $\begingroup$ In your double integral, $r$ changes from $0$ to $\sqrt {a^2 - b^2}$ and the height is taken to be $\sqrt {a^2 - r^2}$. This gives the intersection of the upper hemisphere with the cylinder of radius $\sqrt {a^2 - b^2}$ centered on the $z$ axis, not what you want. Besides, the value of that integral is $2 \pi (a^3 - b^3)/3$, different from your result. The volume of $B$ as a body of revolution is $$\pi \int_b^a (a^2 - x^2) dx,$$ the book formula is correct. $\endgroup$ – Maxim Aug 23 '18 at 16:29
0
$\begingroup$

First determine the volume you integrate over. The region in the $x,y$ plane is bounded by $x^2+y^2=a^2$. The height $z-$coordinate is bounded from above by $z=b$ and from below by $z=\sqrt{a^2-x^2-y^2}$. Now it is wise to switch to cylinder coordinates to calculate the final integral.

$$ V=\int_{0}^a\int_{0}^{2\pi}\int_{\sqrt{a^2-r^2}}^{b}r\,dzd\theta dr $$

$\endgroup$
  • $\begingroup$ But can you do this with a double integral with polar coordinates? On some questions it is more obvious for me to use the double integral and so I dont know why my above way of calculating it is wrong, since this is just another volume under the curve problem which i generally use double integrals for, but here the region is just the circle I defined above $\endgroup$ – NoteBook May 27 '18 at 14:05
  • $\begingroup$ What did you want to be the function to integrate over? For a double integral, you can get the volume under the graph of some function above the $x,y$ plane. In this case, we want to know the volume between the ball and the plane $z=b$, so it is harder to come up with a function you can integrate to get the desired volume $\endgroup$ – user408856 May 27 '18 at 14:07
  • $\begingroup$ So you're saying my method works only when the 'circle' is on the xy plane? If so theres another question in the book where again they use cylindrical coordinates to find the volume under the cone $z=4-\sqrt{x^2+y^2}$ bounded by the xy plane and I use polar coordinates- double integral and again its a different answer $\endgroup$ – NoteBook May 27 '18 at 14:12
  • $\begingroup$ That should result in the same answer, but I don't know your progress. Is your book Adams and Essex calculus? Then please tell me the exercise number $\endgroup$ – user408856 May 27 '18 at 14:14
  • $\begingroup$ Well im looking at the online document given to us which we refer as 'the book'. The solutions are given but dont explain why something is done. mathstat.concordia.ca/faculty/cdavid/EMAT212/solintegrals.pdf Problems 16 for the sphere and 18 for the cone $\endgroup$ – NoteBook May 27 '18 at 14:17
0
$\begingroup$

What was this "different answer" your book has? Did you realize that your answer is equivalent to $\frac{\pi}{2}(a^4- b^4)$?

$\endgroup$
  • $\begingroup$ Yes, the book has answer with $a^3$ and $b^3$ which cant be derived from my answer $\endgroup$ – NoteBook May 27 '18 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.