Volume with cross sections: semicircle | AP Calculus AB | Khan Academy

The volume of a solid with semi-circular cross sections and a triangular base.
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[@TWPO]

Thank you, Sal. This was destroying my brain in class but you described it flawlessly

[@rabidpinkbunny8915]

THANK YOU. I feel like I understand this a lot better from watching this video than by staring at my textbook. XD

[@chauphan1379]

I LIKE THIS VERY GOOD TEACHER. HIS WAY OF TEACHING IS VERY CLEAR.

[@IshowFUNNYvids]

God damn, I can't wait to learn Calculus!

[@danfow22]

Was stuck on a Calc problem like normal. Thanks again Khan academy.

[@electrious_46]

Bro saving my life in uni

[@TheKvacKaj]

Hello, I am not used to use integrals, but wouldn't it be easier in this case to solve it like this:
We know that the radius of the base is 0,5 and we also know that it's a semi cone. We can therefore apply volume of a cone formula divided by two: (pi*r^2*h)/6 (r=0,5;h=1) and the result is pi/24 as we get from solving definite integral.

[@jadriantan9917]

Amazing explanation.

[@Lucuskane]

Move the constants, its not hard, but I have to remember that too often

[@egelsia7009]

I'm facing with the same problem now, hope you have overcome that in past 6 years.

[@YAHADVENTURES90s]

still benefiting from this

[@ionjerdea8147]

SUPER...

[@marielle6673]

Thank you Khan😭

[@amruthagujjar]

thank you!!

[@cursodedibujo-maritoaranda2980]

Love it!

[@polaw7204]

Refreshing

[@gary3ward]

Sorry, meant to say 1-x, my point being that is the radius, not the diameter.

[@bluegrovyle5480]

That is not necessarily true. The problem will give you the information required to determine the value of the radius. Perhaps it could have been, but then the cross sections would naturally be quarter circles and not semi circles.

[@TheJaguarthChannel]

I am sure that this is an extremely misshapen cone, and it breaks my brain when I try to fit the perfect half-cone he drew onto the graph he drew. Like, in side of the base of the cone must extend further than the other, due to the shape of the triangle that denoted the cross-section.

[@gary3ward]

x-1 looks like the radius of the half circle, not the diameter. please correct me if I'm wrong. The volume of a cone formula, V=(1/3)x pi x r^2 x h, so cone with radius 1 and height of 1 has a volume of pi/3. The half cone volume would be pi/6. Please correct me if I am wrong.

[@samali108]

Gary, I was thinking the same way first. However looking at his sketch again I noticed the x and y axis as shown would indicate that the solid figure is laying on its side along the x axis and the large diameter is laying along the y axis. It then made sense of what he is doing.

[@gary3ward]

Shyamal Basu You are correct. I am used to looking at figures generated by rotating around the x-axis. So what he really drew is a skewed half-cone.

[@samali108]

Yes, I believe that's exactly what he has drawn.

[@aindoria]

You made this extra confusing for me because you added a perpendicular that didn't originally exist...

[@TheKglennon]

Couldn't you solve this by using disk method and the Cavalieri theorem?

[@PerfectlyCrafted]

Kyle Glennon nope

[@john-zf1yb]

@@PerfectlyCrafted why not?

[@Aaron-pm7ml]

First