Mass of Region Bounded by y=sin(x), z=1-y, z=0, and x=0

[TeslaCoil137]

Homework Statement

On a sample midterm for my Calc 3 class the following question appears:

Find the mass of (and sketch) the region E with density $\rho = ky$ bounded by the 'cylinder' $y =\sin x$ and the planes $z=1-y, z=0, x=0$ for $0\le x\le\pi/2$.

Homework Equations

$$ m= \int_{E} \rho dV$$

The Attempt at a Solution

By projecting down to the x-y plane the region is given by $(x,y,z): 0\le x\le\pi/2, \sin x\le y\le1, 0\le z\le1-y$.

So the mass of the region is $$\int_{0}^{\pi/2} \int_{\sin x}^1 \int_{0}^{1-y} ky dzdydx$$.

Performing all the iterated integrations gives $k(\frac{2}{9} -\frac{\pi}{24})$ but the stated answer is $k(\frac{16-\pi}{72})$.

 

[Mark44]

Homework Statement

On a sample midterm for my Calc 3 class the following question appears:

Find the mass of (and sketch) the region E with density $\rho = ky$ bounded by the 'cylinder' $y =\sin x$ and the planes $z=1-y, z=0, x=0$ for $0\le x\le\pi/2$.

Homework Equations

$$ m= \int_{E} \rho dV$$

The Attempt at a Solution

By projecting down to the x-y plane the region is given by $(x,y,z): 0\le x\le\pi/2, \sin x\le y\le1, 0\le z\le1-y$.

So the mass of the region is $$\int_{0}^{\pi/2} \int_{\sin x}^1 \int_{0}^{1-y} ky dzdydx$$.

Performing all the iterated integrations gives $k(\frac{2}{9} -\frac{\pi}{24})$ but the stated answer is $k(\frac{16-\pi}{72})$.

The two expressions are equal. Combine the two terms in the first expression using a common denominator of 72.

 

[TeslaCoil137]

Oh, well then my terrible algebra skills strike again. Thanks