15.5: Surface Area

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Alexandra Niedden

Alexandra Niedden

Күн бұрын

Пікірлер: 43
@daniloenriquealvaradoherna403
@daniloenriquealvaradoherna403 Жыл бұрын
Hello, can you explain, why did you put down x when you are in dy, the exercise number 1 of the video.
@alexandraniedden5337
@alexandraniedden5337 Жыл бұрын
I put a few steps together rather than showing my work. When you complete the first integration, you should end up with y*sqrt(4x^2 + 5). Then using those limits of integration from 0 to x for y, this substitution gives x*sqrt(4x^2 + 5).
@danishdar2718
@danishdar2718 4 жыл бұрын
Vector normal to the surface z=1-x^2-y^2, Z>=0 will be
@ruwaqatmash8543
@ruwaqatmash8543 7 ай бұрын
for example two of surface area why is the boud of the polar intergral not from -3 to 3 instead of 0 to 3 at 8:54
@alexandraniedden5337
@alexandraniedden5337 7 ай бұрын
We consider r to be positive, hence our radius extends from 0 (the origin) to 3.
@RAJ_IITIAN1
@RAJ_IITIAN1 Жыл бұрын
Excellent explanation ❤
@muzzammilyousuf2265
@muzzammilyousuf2265 7 ай бұрын
Which book do you use to find the questions or from where? it'll be a great help to practice more
@alexandraniedden5337
@alexandraniedden5337 7 ай бұрын
I use Stewart's Multivariable Calculus (8th edition) :) I have previously used Anton-Bivens-Davis Multivariable Calculus, which is a great resource as well.
@muzzammilyousuf2265
@muzzammilyousuf2265 7 ай бұрын
ok thanks alot
@ahmedmagdyelmedany1678
@ahmedmagdyelmedany1678 2 жыл бұрын
Can we find the surface area for ellipsoid X2/a2+Y2/b2+Z2/c2=1
@hidayatsiddiqui3074
@hidayatsiddiqui3074 4 жыл бұрын
You...did in very shortcut madam..., Where all those steps gone
@jamesgersbach1223
@jamesgersbach1223 4 жыл бұрын
Stop complaining
@mmq6525
@mmq6525 Жыл бұрын
The steps she skips are trivial and already understood by her students
@joshuageorge7997
@joshuageorge7997 3 жыл бұрын
For the definite integral with U-Sub, wouldn't you have to change the bounds for the integral?
@greensparkles6796
@greensparkles6796 Жыл бұрын
You could and keep the value as u^3/2 from 5 - 9, or keep the expression in terms of X and keep the bounds 0 - 1. You will ultimately end up with the same thing.
@minguyen6743
@minguyen6743 3 жыл бұрын
your examples are excellent. i have a question to ask though. i don't know how to present a point in the three coordinate planes. so when you draw the figure for z=9 i get confused. i know i need to know how to draw the objects to find the interval of r as well as theta angle - yet i have no clue to draw when it relates to z-plane. can you please give me some sources (youtuve video, pdf files,...) that teach how to do that? i have searched in the internet ever since last year but still don't get the answer. since you draw it so smoothly i would like to ask you for a favour. thank you
@alexandraniedden5337
@alexandraniedden5337 3 жыл бұрын
Hello! Here is a video where I teach graphing quadric surfaces: kzbin.info/www/bejne/foWWY5yZfayfbM0. This may help with graphing the complicated figures. Good luck!
@kshitijmehta3959
@kshitijmehta3959 3 жыл бұрын
Regards mam, i really love your videos and explanation. Can you please also make a video for 15.8
@johnsears5350
@johnsears5350 4 жыл бұрын
Is it possible to analytically solve the surface area of a elliptic paraboloid? An example would be z=x^2+4y^2, or does a have to equal b in the formula for the shape to be solved by hand.
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
a does not have to equal b. You can still use the formula.
@عطاالدهامشة-ع8ح
@عطاالدهامشة-ع8ح 3 жыл бұрын
Find the surface area due to the rotation of the area between f(x)=x^3 and g(x)=x
@bruceallenbieren5138
@bruceallenbieren5138 4 жыл бұрын
is your formula applicable to all surface area for parabolloid?
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
Yes!
@danavanesa9630
@danavanesa9630 4 жыл бұрын
Why do you use 3 as the limit?
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
Are you referring to example 2? Around 8:40? Assuming so, I switched from rectangular to polar coordinates. The inner limits of integration refer to the radius, r, of the circle. The upper circle is x^2 + y^2 = 9, so the radius ranges from 0 (at the origin) to 3 (at z = 9).
@danavanesa9630
@danavanesa9630 4 жыл бұрын
@@alexandraniedden5337 So instead if the height of the paraboloid was b and the radius was a, then the limit would be a or root square of b?
@danavanesa9630
@danavanesa9630 4 жыл бұрын
this meaning z=b and the radius = a
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
@@danavanesa9630 In the case you are describing, a = sqrt(b), so the upper limit of integration could be a or sqrt(b).
@hessahalamhouj5588
@hessahalamhouj5588 2 жыл бұрын
how did you get y=x in ex#1?
@mmq6525
@mmq6525 2 жыл бұрын
point 0,0 to 1,1. you can do slope formula to get m and just y-y1 = m(x-x1) to get the formula, but it's pretty obvious because it goes from 0,0 to 1,1 and is a straight line that the equation is y=x
@gouthamyadav3021
@gouthamyadav3021 10 ай бұрын
hi at last two steps explain ones video time is 6:00
@mominazhar7275
@mominazhar7275 4 жыл бұрын
Which book youre following ?
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
We use Stewart's Calculus - 8th edition.
@mominazhar7275
@mominazhar7275 4 жыл бұрын
@@alexandraniedden5337 you're great thanks, i wonder if i can ask a question too
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
@@mominazhar7275 You can definitely ask a question. I do not promise that I'll know the answer though.
@mominazhar7275
@mominazhar7275 4 жыл бұрын
@@alexandraniedden5337 i just concerned that how did you conclude to (4r^2 + 1)^(1/2)
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
@@mominazhar7275 I switched from rectangular to polar coordinates. Remember that x^2 + y^2 = r^2.
@Swish-Stories
@Swish-Stories 4 жыл бұрын
I dont understand how you got 4pi/3 in the end. After integrating 2 dtheta from 0 to 1, I got 2. Then integrating that I ended up with 8. What am I doing wrong?
@alexandraniedden5337
@alexandraniedden5337 4 жыл бұрын
We need to change the limits of integration - the 0 to 1 are the limits for x, not theta. Substituting 0 and 1 into x = 2sin(theta) and solving for theta, we get theta = 0 and pi/6. The outer limits of integration are okay, but the inner should be 0 to pi/6. I apologize for the poor notation!
@Swish-Stories
@Swish-Stories 4 жыл бұрын
@@alexandraniedden5337 Thank You!
@Swish-Stories
@Swish-Stories 4 жыл бұрын
@@alexandraniedden5337 Also do you have a video of 15.8 or if you know of a video that has the same content?
@shivamsharma-wm1ob
@shivamsharma-wm1ob 3 жыл бұрын
nectar to deceased
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