Double Integrals: cartesian -> polar and solve

• GreenLantern
In summary, GL19 is stuck on a problem involving double integrals and spheres. He has found r\leq \sqrt{1/2} and z\leq r but is missing information about what to do next. He is able to solve the problem by finding the minimum and maximum values for r and then integrating only r. He also forgot to apply a restriction on R.
GreenLantern
Double Integrals: cartesian --> polar and solve

here is everything:

#19: I am stuck...This is to be solved using cylindrical polar coordinates and a double integral. I understand simpler ones such as find the volume of the solid under the cone z= sqrt(x^2 + Y^2) and above the disk (x^2 + y^2 ) <= 4 but i guess what is throwing me on this one is the sphere and that it is asking for area above the cone...

#13: What did i do wrong? My answer is similar to the book but not quite...

Thanks everyone!

-GL

Last edited:

19:
You have very nearly solved this one. You have found $z \geq r$ and $z \leq \sqrt{1-r^2}$, so all you need to do is find the minimum and maximum values for $r$ and you will have the triple integral
$$\int_{\min(r)}^{\max{r}}\int_r^{\sqrt{1-r^2}}\int_0^{2\pi}rd\theta dz dr$$

13:
You forgot to apply the restriction on R that says $y \leq x$

ughhhh okay and how exactly do i find R min/max? i hardly know how i got to where i did to be honest...

im trying to do it by visual inspection which is how my prof. said to do other ones but my mind is turning to mush thinking about it.I'd say that r max would be 1? due to part of the sphere is included and its radius is 1. Is this correct?
and is the minimum r=0 because the point of the cone?
thanks so much for replying to me though!

Actually, looking at your work again, you already found $r\leq \sqrt{1/2}$. Since this is the only restriction on r, the minimum will be 0.

okay. and then why are we not integrating anything other than r?

WOWWWW
okay so totally disregard me as an asshat because earlier i was totally thinking about doing the integral of $$(r)[ \sqrt{1-r^{2}}-r]$$

with R between 0 and $$\sqrt{1/2}$$
and ø between 0 and 2π

as it would make sense to take the volume of the sphere over the cone and subtract the volume of the area under the cone...WOW i suck :(

sorry for wasting space :) and thanks for the crucial help!

-Ben

Do you need help with #13 still?

no sir i figured it out.

Thanks!

1. What is the difference between cartesian and polar coordinates?

Cartesian coordinates use the x and y axes to locate a point on a plane, while polar coordinates use a distance from the origin and an angle to locate a point on a plane.

2. How do you convert from cartesian coordinates to polar coordinates?

To convert from cartesian coordinates (x,y) to polar coordinates (r,θ), you can use the equations r = √(x^2 + y^2) and θ = tan^-1(y/x).

3. What is the purpose of using double integrals in cartesian to polar conversion?

Double integrals are used to find the area or volume between two curves in cartesian coordinates. When converting to polar coordinates, double integrals can also be used to find the area or volume between two polar curves.

4. How do you solve double integrals in polar coordinates?

To solve double integrals in polar coordinates, you first need to determine the limits of integration for r and θ. Then, you can use the polar form of the integrand and integrate with respect to r and θ, using the proper limits.

5. What are some common applications of double integrals in polar coordinates?

Double integrals in polar coordinates are commonly used in physics, engineering, and other sciences to calculate areas, volumes, and moments of inertia. They are also used in various mathematical and statistical models and in computer graphics to represent circular or symmetric objects.

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