{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solns-ps7-2010

# solns-ps7-2010 - 18.02A IAP 2010 Solutions Problem Set 7...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 18.02A IAP 2010: Solutions Problem Set 7 Part II (30 points) Problem 1. We use cylindrical coordinates: x = r cos θ , y = r sin θ , and z , where in our case 0 ≤ z ≤ y/ 2 = ( r sin θ ) / 2, from which we get y ≥ 0, i.e., 0 ≤ θ ≤ π . From x 2 + y 2 ≤ 1 we get r ≤ 1. Then the mass of the wedge W is: ZZZ W p x 2 + y 2 + z 2 dV = ZZZ W r 2 dz dr dθ = Z π Z 1 Z r sin θ 2 r 2 dz dr dθ = Z π Z 1 r 3 sin θ 2 dr dθ = 1 2 Z π sin θ dθ Z 1 r 3 dr = 1 2 [- cos θ ] π θ =0 r 4 4 1 r =0 = 1 4 . Problem 2 We use spherical coordinates, in which we have a ≤ ρ ≤ b . The density being proportional to the center means that δ ( x,y,z ) = α p x 2 + y 2 + z 2 = αρ . Then the mass of the spherical shell D is: ZZZ D δ ( x,y,z ) dV = ZZZ D αρ 3 sin φdρdφdθ = Z 2 π Z π Z b a αρ 3 sin φdρdφdθ = α Z 2 π dθ Z π sin φdφ Z b a ρ 3 dρ = α · 2 π · 2 · b 4 4- a 4 4 = απ ( b 4- a 4 ) . Problem 3 Place the sphere so the point mass P is at the origin. Let Q be diametrically opposite to P (so it has coordinates (0 , , 2 a )), and form the right triangle with vertices P , Q , and R , where R is an arbitrary point on the surface of the sphere. By trigonometry, weis an arbitrary point on the surface of the sphere....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

solns-ps7-2010 - 18.02A IAP 2010 Solutions Problem Set 7...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online