Calculating Mass of Half Disc Plate with Proportional Density | Physics Homework

[chetzread]

Homework Statement

A plate is in the form of half disc of radius a and placed at positive y-axis. Given that the density of plate is directly proportional to the distance of the straight edge of the plate . Find the mass

Homework Equations

The Attempt at a Solution

$$\int_{0}^\pi \int_{0}^a\ kyr \, dr \, d\theta$$

i gt the ans is 2k(a^3)/3 , but the ans is k(a^3)/3

P/s : density is directly proportional to ky [/B]

 

[Jonathan Scott]

You don't seem to have anything in there to represent the fact that ##y## describes a half disc. You could for example define ##y## as a function of ##x##, or convert to polar coordinates and define ##y## as a function of ##r## and ##\theta##.

 

[Mark44]

You don't seem to have anything in there to represent the fact that ##y## describes a half disc. You could for example define ##y## as a function of ##x##, or convert to polar coordinates and define ##y## as a function of ##r## and ##\theta##.

The limits of integration of the polar integral show that the region of integration is a half circle.

Homework Statement

A plate is in the form of half disc of radius a and placed at positive y-axis. Given that the density of plate is directly proportional to the distance of the straight edge of the plate . Find the mass

Homework Equations

The Attempt at a Solution

$$\int_{0}^\pi \int_{0}^a\ kyr \, dr \, d\theta$$

i gt the ans is 2k(a^3)/3 , but the ans is k(a^3)/3

P/s : density is directly proportional to ky

I get the same answer as you do, so either we have both made the same mistake, or there is an error in the book's answer.

In your integral above, you show y in the integrand. I assume that you changed this to ##r\sin \theta## in your work.

 

[chetzread]

The limits of integration of the polar integral show that the region of integration is a half circle.

I get the same answer as you do, so either we have both made the same mistake, or there is an error in the book's answer.

In your integral above, you show y in the integrand. I assume that you changed this to ##r\sin \theta## in your work.

Yes , I use X = R cos theta , and y = r sin theta