Calculus Integrals for Volumes with Known Cross Sections

In this video we talk about how to find volumes with known cross sections using calculus. I highly recommend that you memorize a couple of formulas for areas: squares, equilateral triangles, semicircles, rectangles, and general triangles. After going over the idea we do some examples perpendicular to the x-axis and others perpendicular to the y-axis, which is just a little more challenging.

[@connieguo230]

I love the way you explain everything. It's so easy to understand.

[@blacklightning9649]

watching this before AP Calc test tomorrow morning, thanks!

[@dawning1640]

You dropped this 👑

[@bertievyas80]

You are literally a savior

[@notimportant9263]

This was a pretty awesome video. Thank you!

[@shreyway12]

this video is amazing thank you brother turk

[@nabojeetdas2515]

You are the best 💯💯

[@hannahthornton3024]

king 🙏🏼

[@pranavmadhukar12]

Thanks for the video! Quick question - so the pi/8 for the semi-circle comes only if the f(x)-g(x) is the diameter right? If the question says that the f(x)-g(x) is the radius for the semi-circle, then the volume becomes s^2 * pi/2 right?

[@turksvids]

Yup! (In my experience it is far less likely--for whatever reason--for that to be the radius.)

[@pranavmadhukar12]

@@turksvids alright! Thanks!

[@harmanbans]

socratic.org/questions/how-do-i-find-the-volume-of-this-integral - the pi/8 formula you gave for a cross section with a semicircle didn't work for this problem. I used pi/2 just from logic, and got it right. Where is the pi/8 from?

[@turksvids]

pi/8 shows up in the integral in that solution.
let y1 = (3/2(16-x^4)^(1/4)) and y2 = -(3/2(16-x^4)^(1/4))
volume is: pi/8*integral( (y1-y2)^2 ),
which ends up 9pi/8 integral( sqrt(16-x^4), -2, 2), which is an ugly integral but matches the solution on that page.

[@mr.doubleplus8206]

how do u find the intersection tho

[@angrynooblets34]

Either set the functions equal to each other and solve or graph then on a calculator and find the intersects