Let's break down the problem step by step: ### Problem Overview - **Region:** Bounded between \( y = \sqrt{x} \) and \( y = x^2 \) - **Axis of Rotation:** y-axis (vertical axis) - **Incorrect Integral Setup:** \[ \int_0^1 \pi (x^2)^2 - \pi (\sqrt{x})^2 \, dx \] (i.e., \(\pi \int_0^1 (x^4 - x) dx\)) ### Why is this setup incorrect? #### 1. **Incorrect Variable of Integration** - The region is being rotated about the **y-axis**, but the integral is set up with respect to **\(dx\)**, i.e., horizontal slices. - **Washer method** with respect to \(x\) is **for rotation about the x-axis**, not the y-axis. - For rotation about the **y-axis**, you either: - Rewrite the functions in terms of \(y\) and integrate with respect to \(dy\) (using horizontal washers/disks), **or** - Use the **shell method** with respect to \(dx\). #### 2. **Functions Are Not Solved for \(x\) in Terms of \(y\)** - For washer/disk method about the y-axis, we need \(x\) as a function of \(y\): - \( y = \sqrt{x} \implies x = y^2 \) - \( y = x^2 \implies x = \sqrt{y} \) #### 3. **Correct Integral Setup** - **Washer method about the y-axis:** \[ \text{Volume} = \int_{y=0}^{y=1} \pi \left( [\text{outer radius}]^2 - [\text{inner radius}]^2 \right) \, dy \] Here, - **Outer radius:** From y-axis to \(x = \sqrt{y}\) - **Inner radius:** From y-axis to \(x = y^2\) So, \[ V = \int_{y=0}^{1} \pi [(\sqrt{y})^2 - (y^2)^2] \, dy = \int_0^1 \pi (y - y^4) \, dy \] ### **Brief Explanation for Why the Given Integral is Incorrect** > The integral your friend set up, \(\int_0^1 \pi (x^2)^2 - \pi (\sqrt{x})^2 \, dx\), does **not** give the correct volume because it integrates with respect to \(x\) as if rotating around the x-axis, not the y-axis. For rotation about the y-axis, the functions should be rewritten with \(x\) in terms of \(y\) and integrated with respect to \(y\), or the shell method should be used. Integrating with respect to \(x\) here does not properly account for the geometry of rotation about the y-axis. --- ### **Final Answer Explanation** **The integral \(\int_0^1 \pi (x^2)^2 - \pi (\sqrt{x})^2 \, dx\) is incorrect because it finds the volume of revolution about the x-axis, not the y-axis. To find the volume about the y-axis, you must express \(x\) in terms of \(y\) and integrate with respect to \(y\), or use the shell method. The correct integral by the washer method is:** \[ \int_0^1 \pi \left( (\sqrt{y})^2 - (y^2)^2 \right) dy = \int_0^1 \pi (y - y^4) dy \]