Question

(1 point)
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the answer is not correct.
which of the following integrals represents the area of the surface obtained by rotating the curve ( y=sec (x), 0 leq x leq pi / 4 ), about the ( y )-axis?
(square) a. ( 2 pi int_{0}^{pi / 4} x sqrt{1+(sec (x) \tan (x))^{2}} d x)
(square) b. ( 2 pi int_{0}^{pi / 4} sec (x) sqrt{1+left(\tan ^{2}(x)
ight)^{2}} d x)
(square) c. ( 2 pi int_{0}^{pi / 4} sec (x) sqrt{1+(sec (x) \tan (x))^{2}} d x)
(square) d. ( 2 pi int_{0}^{pi / 4} x sqrt{1+(2 sec (x) \tan (x))^{2}} d x)
(square) e. ( 2 pi int_{0}^{pi / 4} sec (x) sqrt{1+(2 sec (x) \tan (x))^{2}} d x)
(square) f. ( 2 pi int_{0}^{pi / 4} x sqrt{1+left(\tan ^{2}(x)
ight)^{2}} d x)

Explanation:

Step1: Recall the formula for the surface area of revolution about the \(y -\)axis

The formula for the surface area \(S\) of the surface obtained by rotating the curve \(y = f(x)\) from \(x=a\) to \(x = b\) about the \(y-\)axis is \(S=2\pi\int_{a}^{b}x\sqrt{1+(y')^{2}}dx\).
Given \(y=\sec(x)\), then \(y'=\sec(x)\tan(x)\) (using the derivative formula \(\frac{d}{dx}\sec(x)=\sec(x)\tan(x)\)).

Step2: Substitute \(y\) and \(y'\) into the formula

Substitute \(a = 0\), \(b=\frac{\pi}{4}\), \(y=\sec(x)\) and \(y'=\sec(x)\tan(x)\) into \(S = 2\pi\int_{a}^{b}x\sqrt{1+(y')^{2}}dx\).
We get \(S=2\pi\int_{0}^{\frac{\pi}{4}}x\sqrt{1+(\sec(x)\tan(x))^{2}}dx\).

Answer:

A. \(2\pi\int_{0}^{\frac{\pi}{4}}x\sqrt{1+(\sec(x)\tan(x))^{2}}dx\)