Post Reply 
TestLaTex
11-04-2023, 07:02 PM (This post was last modified: 11-04-2023 07:31 PM by Thomas Klemm.)
Post: #9
RE: TestLaTex
Use \tfrac instead of \dfrac in the \sqrt of the lower limit:

\( \dfrac{4}{3} \times \pi \times r^{3} - 2ab \sqrt{r^{2}- \left ( \dfrac{a}{2} \right ) ^2 - \left ( \dfrac{b}{2} \right ) ^2} - 8 \int_{\sqrt{r^{2} - \left ( \tfrac{a}{2} \right ) ^2 - \left ( \tfrac{b}{2} \right ) ^2}}^{r} \;\; \int_{0}^{\sqrt{r^2-y^2}} \sqrt{r^2-y^2-x^2} \;\; \mathrm{d}x \;\; \mathrm{d}y \)

Use [] instead of ():

\[ \dfrac{4}{3} \times \pi \times r^{3} - 2ab \sqrt{r^{2}- \left ( \dfrac{a}{2} \right ) ^2 - \left ( \dfrac{b}{2} \right ) ^2} - 8 \int_{\sqrt{r^{2} - \left ( \tfrac{a}{2} \right ) ^2 - \left ( \tfrac{b}{2} \right ) ^2}}^{r} \;\; \int_{0}^{\sqrt{r^2-y^2}} \sqrt{r^2-y^2-x^2} \;\; \mathrm{d}x \;\; \mathrm{d}y \]

Or then set \displaystyle explicitly:

\(\displaystyle \dfrac{4}{3} \times \pi \times r^{3} - 2ab \sqrt{r^{2}- \left ( \dfrac{a}{2} \right ) ^2 - \left ( \dfrac{b}{2} \right ) ^2} - 8 \int_{\sqrt{r^{2} - \left ( \tfrac{a}{2} \right ) ^2 - \left ( \tfrac{b}{2} \right ) ^2}}^{r} \;\; \int_{0}^{\sqrt{r^2-y^2}} \sqrt{r^2-y^2-x^2} \;\; \mathrm{d}x \;\; \mathrm{d}y \)

Does this look better?
Find all posts by this user
Quote this message in a reply
Post Reply 


Messages In This Thread
TestLaTex - Tugdual - 04-18-2016, 01:42 PM
RE: TestLaTex - jebem - 04-18-2016, 02:19 PM
RE: TestLaTex - Howard Owen - 04-18-2016, 11:45 PM
RE: TestLaTex - Tugdual - 04-19-2016, 07:08 AM
RE: TestLaTex - Katie Wasserman - 04-19-2016, 10:03 AM
RE: TestLaTex - Tugdual - 04-19-2016, 11:27 AM
RE: TestLaTex - pier4r - 03-25-2017, 08:27 AM
RE: TestLaTex - DM48 - 11-04-2023, 04:23 PM
RE: TestLaTex - Thomas Klemm - 11-04-2023 07:02 PM
RE: TestLaTex - Thomas Klemm - 11-04-2023, 07:09 PM
RE: TestLaTex - Thomas Klemm - 11-04-2023, 07:22 PM
RE: TestLaTex - Jonathan Busby - 11-05-2023, 04:00 PM



User(s) browsing this thread: 6 Guest(s)