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Use spherical coordinates to evaluate the triple integral of the function J(J) over the solid, bounded by the surfaces 3 . Kri...

Question

Use spherical coordinates to evaluate the triple integral of the function J(J) over the solid, bounded by the surfaces 3 . Kri

Use spherical coordinates to evaluate the triple integral of the function J(J) over the solid, bounded by the surfaces 3 . Kri



Answers

Use spherical coordinates.

Evaluate $ \iiint_E y^2 z^2\ dV $, where $ E $ lies above the cone $ \phi = \frac{\pi}{3} $ and below the sphere $ \rho = 1 $

So, um, oftentimes we can't just take a single integral. We often have to take double integral Zoran this case triple integral in order to find important values. In this case, we're finding volume. We also know that it's important to be able to switch between coordinates because, um, things aren't always rectangular. Oftentimes it's easier to use cylindrical or spherical coordinates. Eso What we're going to be doing is working on our ability to use those different coordinates to make problems ultimately easier. So what we have here is we first want to determine the range for theta. Um, since we know the regions in the first Occident, the angle will range from zero to pi over two. Then the Z value. We want to convert into rectangular coordinates. Since Z is equal to four minus x squared plus y squared, we will just call this four minus r squared since x minus X squared minus y plus y squared is R squared. Then we know that when Z uh, we can plug in Z equals zero. When we do that, we see that our is equal. Teoh a radius of two. So with all this in mind, we now can take are integral. We know that the fate of value is from zero to pi over two. We know that our radius is from 0 to 2 and then we have our functions E which is represented by four minus r squared. And therefore we have our X plus y plus z Easy. De are de Seda. Um what we can dio with this is, um, move the theta right here. Move the theta, right? Yeah, When we do that, we know that this is just gonna be pi over two. Similarly, we can we want to represent And before we do that actually will want to represent X, y and Z as the appropriate, um, variables. So X we know is going to be our coastline. Fada, Why will be our signed Fada and Z will just be easy, but we want to have it's going to be our DZ drd fada. So when we multiply everything through, we get our Z r squared sine data R squared coastline data evaluate this simplify, But we can also use these calculating tools because they're very useful in finding inner grows quickly s So what? We end up getting is this? And another way we could write it if we salt it out, would be 128 over 15. Class 8/3. Hi. Um, as we see these answers, check out. So we know that we did it correctly, and this is how we will conclude this problem.

Were given an integral and rescue spherical coordinates to evaluate it. The integral is the triple integral over the region. E. G. Of Y squared. T. V. Where E is the solid hemisphere right? I thought it was X squared plus Y squared plus C squared is less than or equal to nine. And why is greater than or equal to 0? Just so it's the if you're looking straight on the positive X axis, the right side of the sphere it's almost me dr the doctor of the train. Basic first We want to rewrite our region in terms of spiritual coordinates. So we have that rho squared is less than equal to nine or that row is less than or equal to three. This is our solid sphere of radius three. Now for why? To be greater than or equal to zero. This means that plugging in spherical coordinates we have rho times the sine of phi times the sine of theta is greater than or equal to zero or more simply. Because why is greater than or equal to zero? We know that we're going to restrict data to be Greater than zero and less than or equal to Pi. As for fi there's no restriction. It goes from 0 to Pi. Thank you. and row goes from 0 to 3 And therefore are integral in spherical coordinates is the integral from 0 to Pi integral. From zero to pi Integral from 0 to 3 of our function Y squared in spherical coordinates. This is now rho squared sine squared five. Sine squared theta. And then D. V becomes rho squared sine phi dear. Oh defy the theta Using few beanies theorem. This is the product of integral integral from zero pi sine squared data. D Theta Times The integral from 0 to Pi uh sign cute fi D five times the integral from 0 to 3. Ah Road at the 4th. Hero taking anti derivatives. Well this is uh negative co sign cubed data over three. Not really. I'm it's a little more complicated. Sorry the anti derivative of sine squared this is uh Data over 2 -1 4 Sign of two. Theta From Theta equals 0 to Pi. Yeah adam can't part true. And an anti derivative of sine cubed fi This is going to be negative. Co sign five Plus co sign Cube defy over three imperial From Phi equals 0 to Pi. Yeah. And he didn't times Road to the 5th over five From row equals 0- three. Whoa. Right. Think plugging in. We get uh 1/2 high or pi over two times four thirds. Yeah. Times 3 to the 5th over five which is equal to 162/5 times pi three. A man died.


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