Question

suppose f(x) > 0 for all x and \\(\int_{0}^{9} f(x) dx = 18\\). let r be the region in the first quadrant bounded by the coordinate axes, \\(y = f(x^2)\\), and \\(x = 9\\). find the volume of the solid generated by revolving r around the y - axis.\
set up the integral that gives the volume of the solid. use increasing limits of integration. select the correct choice below and fill in the answer boxes to complete your choice.\
(type exact answers.)\
\\(\bigcirc\\) a. \\(\int_{\square}^{\square} f(x^2) dx\\)\
\\(\bigcirc\\) b. \\(\int_{\square}^{\square} \left(f(x^2)\
ight)^2 dx\\)

Explanation:

Step1: Identify the method for volume around y - axis

When finding the volume of a solid of revolution around the \(y\) - axis using the disk/washer method (or in this case, since we are rotating around the \(y\) - axis and using the method of cylindrical shells or disk method in terms of \(x\)), if we use the disk method (treating \(x\) as a function of \(y\)) or the shell method. But in the context of the integral setup here, we are using the disk method where the radius of the disk is \(x\) and the height is \(y = f(x^{2})\)? Wait, no, actually, when rotating around the \(y\) - axis, the formula for the volume using the disk method (if we solve for \(x\) in terms of \(y\)) or the shell method. But looking at the options, we are using the disk method where the radius is \(x\) and the area of the disk is \(\pi r^{2}=\pi x^{2}\), but wait, the function is \(y = f(x^{2})\)? Wait, no, the region \(R\) is bounded by the coordinate axes, \(y = f(x^{2})\) and \(x = 9\) in the first quadrant. When rotating around the \(y\) - axis, the volume using the disk method (slicing perpendicular to the \(y\) - axis) would have a radius \(x\) where \(x\) goes from \(0\) to \(9\), and the height of the disk (in the \(x\) - direction) is related to \(y = f(x^{2})\). But actually, the correct formula for the volume when rotating around the \(y\) - axis using the shell method is \(V = 2\pi\int_{a}^{b}x\cdot h(x)dx\), but the options given are in terms of \(\pi\int\) (maybe a typo, and it's supposed to be \(\pi\) or not? Wait, the options are missing the \(\pi\) factor? Wait, no, maybe the original problem has a typo. But looking at the options, option B is \(\pi\int(f(x^{2}))^{2}dx\)? Wait, no, the options are:

Option A: \(\pi\int(f(x^{2}))dx\) (assuming the \(\pi\) is missing, because volume of a disk is \(\pi r^{2}\))

Option B: \(\pi\int(f(x^{2}))^{2}dx\)

Wait, no, actually, when rotating around the \(y\) - axis, if we use the disk method (treating \(x\) as a function of \(y\)), \(x = g(y)\), and the volume is \(V=\pi\int_{c}^{d}(g(y))^{2}dy\). But in our case, we are integrating with respect to \(x\), so we use the shell method. Wait, maybe there is a mis - statement. Wait, the region is bounded by \(x = 0\), \(x = 9\), \(y = 0\) (coordinate axis) and \(y = f(x^{2})\). When rotating around the \(y\) - axis, the radius of the cylindrical shell is \(x\), the height of the shell is \(y = f(x^{2})\), and the thickness is \(dx\). The formula for the volume using the shell method is \(V = 2\pi\int_{a}^{b}x\cdot h(x)dx\), where \(h(x)\) is the height of the shell. But the options given are missing the \(2\pi\) and have \(\pi\) (maybe a mistake). But looking at the options, if we consider the disk method (slicing perpendicular to the \(y\) - axis), the radius \(x\) is a function of \(y\), so \(x=\sqrt{g(y)}\) if \(y = g(x)\). But the function here is \(y = f(x^{2})\), so \(x=\sqrt{f^{- 1}(y)}\) (if \(f\) is invertible). But the options are in terms of \(x\) as the variable of integration.

Wait, maybe the problem has a typo and the function is \(y = f(x)\) instead of \(y = f(x^{2})\), but assuming the function is \(y = f(x)\), when rotating around the \(y\) - axis, the volume using the disk method (slicing perpendicular to the \(y\) - axis) with \(x\) from \(0\) to \(9\), the radius is \(x\) and the height (in terms of \(x\)) is \(y = f(x)\). But no, the correct formula for the volume when rotating around the \(y\) - axis using the disk method (slicing perpendicular to the \(y\) - axis) is \(V=\pi\int_{c}^{d}(x)^{2}dy\), where \(x\) is a function of \(y\). If we use the method of integrating with respect to \(x\) (shell method), \(V = 2\pi\int_{0}^{9}x\cdot f(x^{2})dx\), but the options are \(\int(f(x^{2}))dx\) and \(\int(f(x^{2}))^{2}dx\) (missing \(\pi\) and \(2\pi\) factors). But looking at the options, the correct setup for the volume (assuming a disk method where the radius is \(x\) and the area is \(\pi x^{2}\), but if the height of the disk (in the \(y\) - direction) is \(f(x^{2})\), no, that doesn't make sense. Wait, maybe the function is \(y = f(x)\) and there is a misprint, and the options are supposed to have \(\pi\). But among the two options, when rotating around the \(y\) - axis, the volume formula using the disk method (if we consider the radius as \(x\) and the area of the disk is \(\pi r^{2}=\pi x^{2}\), but the height (in the \(y\) - direction) is \(y = f(x)\). But the options have \(f(x^{2})\). Wait, maybe the region is bounded by \(y = f(x^{2})\), \(x = 0\), \(x = 9\) and \(y = 0\). When rotating around the \(y\) - axis, the radius of the disk (when slicing perpendicular to the \(y\) - axis) is \(x\), and the area of the disk is \(\pi x^{2}\), and the thickness is \(dy\). But to integrate with respect to \(x\), we can use the shell method: \(V = 2\pi\int_{0}^{9}x\cdot f(x^{2})dx\). But the options are missing the \(2\pi\) and have \(\pi\) (maybe a mistake). But among the two options, option B is \(\pi\int(f(x^{2}))^{2}dx\) (if we use the disk method with radius \(f(x^{2})\), which is wrong) and option A is \(\pi\int(f(x^{2}))dx\) (also wrong). Wait, no, maybe the original problem is rotating around the \(x\) - axis? No, the problem says around the \(y\) - axis.

Wait, maybe there is a misinterpretation. Let's re - read the problem: "the region in the first quadrant bounded by the coordinate axes, \(y = f(x^{2})\), and \(x = 9\). Find the volume of the solid generated by revolving \(R\) around the \(y\) - axis".

When rotating around the \(y\) - axis, using the disk method (slicing perpendicular to the \(y\) - axis), the radius of each disk is \(x\), where \(x\) ranges from \(0\) to \(9\), and for a given \(x\), \(y\) ranges from \(0\) to \(f(x^{2})\). To find the volume, we can also use the method of integrating with respect to \(x\) (shell method) or with respect to \(y\) (disk method).

If we use the disk method (integrating with respect to \(y\)):

1. First, solve \(y = f(x^{2})\) for \(x\): \(x=\sqrt{f^{-1}(y)}\) (assuming \(f\) is invertible).

2. The volume \(V=\pi\int_{y = 0}^{y = f(9^{2})}(\sqrt{f^{-1}(y)})^{2}dy=\pi\int_{0}^{f(81)}f^{-1}(y)dy\), which is not in the options.

If we use the shell method (integrating with respect to \(x\)):

1. The radius of the shell is \(x\), the height of the shell is \(y = f(x^{2})\), and the thickness of the shell is \(dx\).

2. The formula for the volume using the shell method is \(V = 2\pi\int_{a}^{b}(\text{radius})\times(\text{height})dx=2\pi\int_{0}^{9}x\cdot f(x^{2})dx\).

But the options given are missing the \(2\pi\) and have \(\pi\) (maybe a typo, and the \(\pi\) is a mistake, or the problem is rotating around the \(x\) - axis). If we rotate around the \(x\) - axis, the volume using the disk method is \(V=\pi\int_{0}^{9}(f(x^{2}))^{2}dx\), which is option B.

Since the problem says "around the \(y\) - axis", but the options are more consistent with rotating around the \(x\) - axis (maybe a misprint in the problem). So assuming that there is a misprint and we are rotating around the \(x\) - axis, the volume using the disk method is \(V=\pi\int_{a}^{b}(y)^{2}dx\), where \(y = f(x^{2})\) and \(x\) goes from \(0\) to \(9\). So the integral is \(\pi\int_{0}^{9}(f(x^{2}))^{2}dx\), which is option B.

Step2: Determine the limits of integration

The region is bounded by \(x = 0\) and \(x = 9\) (since it's bounded by \(x = 9\) and the coordinate axes in the first quadrant), so the limits of integration for \(x\) are from \(0\) to \(9\).

Answer:

The correct integral setup (assuming a misprint and rotating around the \(x\) - axis) is \(\pi\int_{0}^{9}(f(x^{2}))^{2}dx\), so the answer is the option with \(\int_{0}^{9}(f(x^{2}))^{2}dx\) (Option B with limits \(0\) and \(9\)). So the integral is \(\boxed{\pi\int_{0}^{9}(f(x^{2}))^{2}dx}\) (but if we strictly follow the options without \(\pi\) (which is a mistake), the integral is \(\int_{0}^{9}(f(x^{2}))^{2}dx\) with option B).