I guess there is a simple mistake you made in r boundaries. It should be from 0 to 2 not from 0 to r even though you said it is from 0 to 2 in 13:00 . Anyways, I appreciate effort you made in this video. Thanks!
Very beneficial examples to grasp the concept! Just one suggestion, you may use some other perspectives to visualize better, for ex. in the last question, if you look at the solid from the left side of the x-axis, what is the solid looks like can be understood readily. Thank you
Can you help me with this pls? Consider the hot air balloon with equation 9x^2 +9y^2 +4z^2 = 100. The temperature in the hot air balloon is given by the following function: f(x,y,z) = 18x^2 + 18y^2 + 8z^2 − 20 • Convert the regular area into appropriate spherical coordinates. • Calculate the volume of the balloon. • Calculate the average temperature in the Balloon
Should it not be a hemisphere and not a 1/4 of a sphere since z = sqrt(4 - x^2 - y^2) when you square both sides it gives a full sphere also the second integral is bounded by y = sqrt( 4 - x^2) is half a circle not a 1/4 of it
At 12:42 when you're substituting for "r" in spherical coordinates, did you make a typo with the rho being squared. If not can you please explain how you get (p^2 sin(fi) ) when subbing in for "r"
When you switch or transform an integral from one coordinate system to another a Jacobian is needed. The Jacobian comes from transforming x's, y's, and z's (if in 3D) to your new coordinates. For polar it comes from taking a slice of a circle and making that a small rectangle to then be added up by the integration. rho^2sin(phi) is the Jacobian for the transformation from x's, y's and z's to rho's, phi's, and theta's. See the video on transformation of variables: Change of Variables or look at Pauls Online Notes: tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx
Probably doesn't matter to you two years later, but she does start on the positive x axis. If you look at it in terms of the x-z plane, she's starting where x is positive and ends where x is negative. The z just shows where the function is when x is a certain value
hello, the integral with phi from 0 to pi over 3 and -pi half to pi half ; are you sure you get a cone, in my program i don't get the cone/ only what is on top of the cone.
I know it might not be helpful now, but I think we need to take the upper limits of theta and phi as pi by 2 as long the as the area we are conserned with is in a single quadrant. If it includes 2 quadrants, it should be pi, and the lower limit mostly zero.
this video is great! i have been having the worst time trying to draw these objects.. who new not playing video games as a kid would actually end up hurting me instead of helping. -___-
I'm not sure where in the video you are referring to. There is a spot where I substitute 4-r^2 for 4-x^2-y^2. This uses the fact that x^2+y^2=r^2. If this is not the spot then please let me know when in the video this substitution happened.
Thank you! The coordinate systems video was golden! Exactly what I needed!
🌸
I guess there is a simple mistake you made in r boundaries. It should be from 0 to 2 not from 0 to r even though you said it is from 0 to 2 in 13:00 . Anyways, I appreciate effort you made in this video. Thanks!
Very beneficial examples to grasp the concept! Just one suggestion, you may use some other perspectives to visualize better, for ex. in the last question, if you look at the solid from the left side of the x-axis, what is the solid looks like can be understood readily.
Thank you
Can you help me with this pls?
Consider the hot air balloon with equation 9x^2 +9y^2 +4z^2 = 100.
The temperature in the hot air balloon is given by the following function: f(x,y,z) = 18x^2 + 18y^2 + 8z^2 − 20
• Convert the regular area into appropriate spherical coordinates.
• Calculate the volume of the balloon.
• Calculate the average temperature in the Balloon
Should it not be a hemisphere and not a 1/4 of a sphere since z = sqrt(4 - x^2 - y^2) when you square both sides it gives a full sphere also the second integral is bounded by y = sqrt( 4 - x^2) is half a circle not a 1/4 of it
such a short yet helpful video!
i think you have a typo at 13:01 r should be "2"? am I correct. Great video tho! helped alot
I noticed this as well
Confused the fuck out of me
Thank you, you're a great teacher.
At 12:42 when you're substituting for "r" in spherical coordinates, did you make a typo with the rho being squared. If not can you please explain how you get (p^2 sin(fi) ) when subbing in for "r"
When you switch or transform an integral from one coordinate system to another a Jacobian is needed. The Jacobian comes from transforming x's, y's, and z's (if in 3D) to your new coordinates. For polar it comes from taking a slice of a circle and making that a small rectangle to then be added up by the integration. rho^2sin(phi) is the Jacobian for the transformation from x's, y's and z's to rho's, phi's, and theta's. See the video on transformation of variables: Change of Variables or look at Pauls Online Notes: tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx
On 4.25, for theta, can you explain why you said it will start on the x axis, but you started on the z axis instead?
Probably doesn't matter to you two years later, but she does start on the positive x axis. If you look at it in terms of the x-z plane, she's starting where x is positive and ends where x is negative. The z just shows where the function is when x is a certain value
@@mdrrolling6582 it matters to me. thanks
hello, the integral with phi from 0 to pi over 3 and -pi half to pi half ; are you sure you get a cone, in my program i don't get the cone/ only what is on top of the cone.
But how do you set the iterated integral ?
Please explain in detail how to get the limits of theta and phi
I know it might not be helpful now, but I think we need to take the upper limits of theta and phi as pi by 2 as long the as the area we are conserned with is in a single quadrant. If it includes 2 quadrants, it should be pi, and the lower limit mostly zero.
this video is great! i have been having the worst time trying to draw these objects.. who new not playing video games as a kid would actually end up hurting me instead of helping. -___-
x^2+y^2=r^2 ,,,,, you had x^2-y^2 yet you replaced it by r^2 , please explain
I'm not sure where in the video you are referring to. There is a spot where I substitute 4-r^2 for 4-x^2-y^2. This uses the fact that x^2+y^2=r^2. If this is not the spot then please let me know when in the video this substitution happened.
12:10
its 4-x^2-y^2 so you can also write it like this 4-(x^2+y^2) which then you can see that you can replace the x^2+y^2 by the r^2
Thnks! It really helped me :)
Nice presentation
Very helpful. Thanks!
Thank you so much
Thank you so much!
Thank you!
Thanks
thanks alot .
too short