Part 1
[tex]x = g(y), \ c \le y \le d\\\\\\x = 3+4y^2, \ 1 \le y \le 2\\\\\\\frac{dx}{dy} = 8y, \ 1 \le y \le 2\\\\\\\displaystyle A = 2\pi \int_{c}^{d}y\sqrt{1+[g'(y)]^2}dy\\\\\\\displaystyle A = 2\pi \int_{c}^{d}y\sqrt{1+\left[\frac{dx}{dy}\right]^2}dy\\\\\\\displaystyle A = 2\pi \int_{1}^{2}y\sqrt{1+(8y)^2}dy\\\\\\\displaystyle A = 2\pi \int_{1}^{2}y\sqrt{1+64y^2}dy\\\\\\\displaystyle A = 2\pi \int_{1}^{2}\sqrt{1+64y^2}*ydy\\\\\\\displaystyle A = 2\pi \int_{65}^{257}\sqrt{u}*\frac{du}{128}\\\\\\[/tex]
In the last two steps I let u = 1+64y^2, so du/dy = 128y which leads to ydy = du/128.
Note the change in the limits of integration. If y = 1, then u = 1+64y^2 = 65. If y = 2, then 1+64y^2 = 257
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Part 2
[tex]\displaystyle A = 2\pi \int_{65}^{257}\sqrt{u}*\frac{du}{128}\\\\\\\displaystyle A = 2\pi*\frac{1}{128}\int_{65}^{257}u^{1/2}*du\\\\\\\displaystyle A = \frac{\pi}{64}\int_{65}^{257}u^{1/2}*du\\\\\\\displaystyle A = \frac{\pi}{64}\left(\frac{2}{3}u^{3/2}+C\right)\Bigg]_{65}^{257}\\\\\\\displaystyle A = \frac{\pi}{64}\left(\left(\frac{2}{3}(257)^{3/2}+C\right)-\left(\frac{2}{3}(65)^{3/2}+C\right)\right)\\\\\\\displaystyle A = \frac{\pi}{64}*\frac{2}{3}\left((257)^{3/2}-(65)^{3/2}\right)\\\\\\[/tex]
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Part 3
[tex]\displaystyle A = \frac{\pi}{96}\left(\sqrt{257^3}-\sqrt{65^3}\right)\\\\\\\displaystyle A = \frac{\pi}{96}\left(\sqrt{257^2*257}-\sqrt{65^2*65}\right)\\\\\\\displaystyle A = \frac{\pi}{96}\left(\sqrt{257^2}*\sqrt{257}-\sqrt{65^2}*\sqrt{65}\right)\\\\\\\displaystyle A = \frac{\pi}{96}\left(257\sqrt{257}-65\sqrt{65}\right)\\\\\\[/tex]
[tex]\displaystyle A = \frac{\pi}{96}*257\sqrt{257}-\frac{\pi}{96}*65\sqrt{65}\\\\\\\displaystyle A = \frac{257\pi\sqrt{257}}{96}-\frac{65\pi\sqrt{65}}{96}\\\\\\\displaystyle A = \frac{-65\sqrt{65}\pi}{96}+\frac{257\sqrt{257}\pi}{96}\\\\\\[/tex]
