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vovikov84 [41]
1 year ago
13

Let R be the region bounded by

Mathematics
1 answer:
loris [4]1 year ago
7 0

a. The area of R is given by the integral

\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36

b. Use the shell method. Revolving R about the x-axis generates shells with height h=x+6-7\sin\left(\frac{\pi x}2\right) when 1\le x\le 2, and h=x+6-7(x-2)^2 when 2\le x\le\frac{22}7. With radius r=x, each shell of thickness \Delta x contributes a volume of 2\pi r h \Delta x, so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56

c. Use the washer method. Revolving R about the y-axis generates washers with outer radius r_{\rm out} = x+6, and inner radius r_{\rm in}=7\sin\left(\frac{\pi x}2\right) if 1\le x\le2 or r_{\rm in} = 7(x-2)^2 if 2\le x\le\frac{22}7. With thickness \Delta x, each washer has volume \pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x. As more and thinner washers get involved, the total volume converges to

\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16<em />

d. The side length of each square cross section is s=x+6 - 7\sin\left(\frac{\pi x}2\right) when 1\le x\le2, and s=x+6-7(x-2)^2 when 2\le x\le\frac{22}7. With thickness \Delta x, each cross section contributes a volume of s^2 \Delta x. More and thinner sections lead to a total volume of

\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70

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