Sun K.

asked • 05/04/13

Find the mass of the plate with density?

Find the mass of the plate with density p(x, y)=ky, where k is a positive constant and R is the region bounded by y=x^2 and x=y^2.

Sun K.

Where did you get 0 to 1?

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05/04/13

Brad M.

tutor

Hey Sun!  Visualize a 1x1 box. The region of overlapping "U-curves" or parabolas is like a pistachio touching the origin at the "southwest" corner running to the NW corner at (1,1). The overlap in so thin that it's a sliver of the box area.

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05/04/13

Grigori S.

This is the explanation why these two curves have two points of  intersection (0,0) and (1,1).

If   y = x^2 and x = y^2 then x^2 = y^4 and we can substitute x^2 into the first equation. We will obtain

                                     y = y^4    or y^4 - y = 0

We can factor this equation and write:

                                 y(y^3 -1)= y(y-1)(y^2 +y+1) = 0

First two factors give us y = 0 and y = 1 as solutions of the equation. Thus x = 0 if y=0 and x = 1 if y=1.

This is visible from graphs of the functions. The third term y^2 +y +1 has only

complex roots and can't be used here.

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05/05/13

Grigori S.

This is the explanation why these two curves have two points of  intersection (0,0) and (1,1).

If   y = x^2 and x = y^2 then x^2 = y^4 and we can substitute x^2 into the first equation. We will obtain

                                     y = y^4    or y^4 - y = 0

We can factor this equation and write:

                                 y(y^3 -1)= y(y-1)(y^2 +y+1) = 0

First two factors give us y = 0 and y = 1 as solutions of the equation. Thus x = 0 if y=0 and x = 1 if y=1.

This is visible from graphs of the functions. The third term y^2 +y +1 has only

complex roots and can't be used here.

Report

05/05/13

2 Answers By Expert Tutors

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Grigori S. answered • 05/04/13

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Certified Physics and Math Teacher G.S.

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The region R is bounded by two curves y=x^2 and y = sqrt(x). They intersect at two points x=0, y=0 , and x=1, y = 1. Mass is equal to the integral of p(x,y) dxdy = int kydxdy. limits of integration by x are variable:

x = y^2 for lower limit and x = sqrt(y) for upper limit (since 0<y<1, y^2 < sqrt(y)).. Integration by x gives us

                         mass = int ky(sqrt(y) - y^2)dy = 3k/20 = 0.15k.

The function p(x,y) represents the surface density, and these calculations are true for infinitely thin plate only.

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Robert J. answered • 05/04/13

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Certified High School AP Calculus and Physics Teacher

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Since the density is the function of y, it is better to do integration over dy.

mass = density x area

So, the mass of the plate = ∫[0, 1] ky (√y - y2)dy = 3k/20 <==Answer

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Robert J.

For your additional question:

The intersection of y=x^2 and x=y^2 is at x = 0, and 1. So, the lower limit is 0, and the upper limit is 1.

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05/06/13

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