Find the mass of the lamina described by the inequalities $0 \leq x \leq 4 \text { and } 0 \leq y \leq 8 \text {, }$ given that its density is $\rho ( x , y ) = 2 x y$. (Hint: Some of the integrals are simpler in polar coordinates.)
A) 768
B) 128
C) 512
D) 256
E) 392

Question

Quizplus · Accepted Answer

We will use double integrals to find the mass of the lamina. We can start by setting up the integral as follows:

$$\iint\limits_{R} 2xy\,dA$$

where $R$ is the region in the $xy$-plane described by the inequalities $0 \leq x \leq 4$ and $0 \leq y \leq 8$. Integrating with respect to $y$ first, we get:

$$\int_{0}^{4} \int_{0}^{8} 2xy\,dy\,dx = \int_{0}^{4} \left[x y^2 ight]_{0}^{8} dx = \int_{0}^{4} 64x dx = 256$$

Therefore, the mass of the lamina is 256. The answer is (D).

However, we can simplify the integral by changing to polar coordinates. The region $R$ is defined by $0 \leq r \leq 8$ and $0 \leq 	heta \leq \frac{\pi}{2}$ in polar coordinates. The integrand becomes:

$$2xy = 2(r \cos 	heta)(r \sin 	heta) = 2r^2\cos 	heta \sin 	heta$$

The double integral becomes:

$$\int_{0}^{\frac{\pi}{2}} \int_{0}^{8} 2r^2\cos 	heta \sin 	heta\,r\,dr\,d	heta$$

Simplifying, we get:

$$\left(\int_{0}^{8} 2r^3\,dright) \left(\int_{0}^{\frac{\pi}{2}} \cos 	heta \sin 	heta\,d	hetaight) = \left(\frac{1}{2} \cdot 2^4 \cdot 8ight) \cdot \frac{1}{2} = 64$$

Therefore, the mass of the lamina is 64. The answer is (B).

Find the Mass of the Lamina Described by the Inequalities 0≤x≤4 and 0≤y≤8, 0 \leq x \leq 4 \text { and } 0 \leq y \leq 8 \text {, }0≤x≤4 and 0≤y≤8,

Find the Mass of the Lamina Described by the Inequalities $0 \leq x \leq 4 \text { and } 0 \leq y \leq 8 \text {, }$