Let's solve **Problem 1** step-by-step using the **method of cylindrical shells**. --- ### **Given:** - \( y = \sqrt{x} \) - \( x = 2y \) - About \( x = 6 \) --- ### **Step 1: Sketch the Region** First, let's find the intersection points: 1. \( y = \sqrt{x} \) 2. \( x = 2y \implies y = \frac{x}{2} \) Set equal: \[ \sqrt{x} = \frac{x}{2} \] \[ 2\sqrt{x} = x \] \[ x^2 - 4x = 0 \implies x = 0 \text{ or } x = 4 \] So, the **limits** are from \( x = 0 \) to \( x = 4 \). --- ### **Step 2: Describe the Region** - **Left boundary:** \( x = 2y \) ⇒ \( y = \frac{x}{2} \) - **Upper boundary:** \( y = \sqrt{x} \) - **Right boundary:** \( x = 4 \) (since at \( x = 4 \), \( y = 2 \)) --- ### **Step 3: Set Up the Shell Method** We are rotating about \( x = 6 \). For a vertical slice at \( x \), the **height** is: \[ \text{height} = \text{upper } y - \text{lower } y = \sqrt{x} - \frac{x}{2} \] The **radius** from \( x \) to \( x = 6 \) is \( 6 - x \). The **volume element** is: \[ dV = 2\pi (\text{radius}) (\text{height}) \, dx = 2\pi (6-x)\left[\sqrt{x} - \frac{x}{2}\right] dx \] Limits: \( x = 0 \) to \( x = 4 \). --- ### **Step 4: Write the Integral** \[ V = 2\pi \int_{0}^{4} (6-x)\left[\sqrt{x} - \frac{x}{2}\right] dx \] Expand: \[ = 2\pi \int_{0}^{4} \left[ (6-x)\sqrt{x} - (6-x)\frac{x}{2} \right] dx \] \[ = 2\pi \int_{0}^{4} \left[ 6\sqrt{x} - x\sqrt{x} - 3x + \frac{x^2}{2} \right] dx \] --- ### **Step 5: Integrate** - \(\int 6\sqrt{x} dx = 6 \cdot \frac{2}{3} x^{3/2} = 4x^{3/2}\) - \(\int x\sqrt{x} dx = \int x^{3/2} dx = \frac{2}{5} x^{5/2}\) - \(\int 3x dx = \frac{3}{2} x^2\) - \(\int \frac{x^2}{2} dx = \frac{1}{2} \cdot \frac{x^3}{3} = \frac{x^3}{6}\) So, \[ V = 2\pi \left[ 4x^{3/2} - \frac{2}{5}x^{5/2} - \frac{3}{2}x^2 + \frac{1}{6}x^3 \right]_{0}^{4} \] --- ### **Step 6: Plug in the Limits** Calculate at \( x = 4 \): - \( x^{3/2} = (4)^{3/2} = 8 \) - \( x^{5/2} = (4)^{5/2} = 32 \) - \( x^2 = 16 \) - \( x^3 = 64 \) So, \[ V = 2\pi \left[ 4(8) - \frac{2}{5}(32) - \frac{3}{2}(16) + \frac{1}{6}(64) \right] \] \[ = 2\pi \left[ 32 - \frac{64}{5} - 24 + \frac{32}{3} \right] \] Find a common denominator (15): - \( 32 = \frac{480}{15} \) - \( \frac{64}{5} = \frac{192}{15} \) - \( 24 = \frac{360}{15} \) - \( \frac{32}{3} = \frac{160}{15} \) So, \[ = 2\pi \left[ \frac{480 - 192 - 360 + 160}{15} \right] \] \[ = 2\pi \left[ \frac{480 + 160 - 192 - 360}{15} \right] \] \[ = 2\pi \left[ \frac{640 - 552}{15} \right] \] \[ = 2\pi \left[ \frac{88}{15} \right] \] \[ = \frac{176\pi}{15} \] --- ## **Final Answer:** \[ \boxed{V = \frac{176\pi}{15}} \] --- **Don't forget to sketch the region and shade the area as required in the instructions.** --- Let me know if you want the step-by-step for problem 2 as well!

Certainly! Please provide the details or image of part 2 of the problem, and I will assist you with a detailed step-by-step solution.

Let's carefully analyze the problem: **Given curves:** - \( x = 2y^2 \) - \( x = y^2 + 1 \) **Rotation axis:** - About \( y = -2 \) --- ### **Step 1: Sketch the region** First, find the intersection points of the two curves: Set equal: \[ 2y^2 = y^2 + 1 \] \[ 2y^2 - y^2 = 1 \] \[ y^2 = 1 \] \[ y = \pm 1 \] Find corresponding \( x \) values: - For \( y = 1 \): - \( x = 2(1)^2 = 2 \) - \( x = 1^2 + 1 = 2 \) - For \( y = -1 \): - \( x = 2(1)^2 = 2 \) - \( x = (-1)^2 + 1 = 2 \) So, the curves intersect at points: \[ (2, 1) \quad \text{and} \quad (2, -1) \] The region is bounded between \( y = -1 \) and \( y = 1 \), and between the curves \( x = 2y^2 \) (left) and \( x = y^2 + 1 \) (right). --- ### **Step 2: Describe the region** - For \( y \) between \(-1\) and \(1\), - The left boundary is \( x = 2y^2 \), - The right boundary is \( x = y^2 + 1 \). --- ### **Step 3: Set up the volume integral (using the shell method)** Since the axis of rotation is \( y = -2 \), which is **horizontal**, the shell method involves **horizontal shells**: - For a given \( y \), the **radius** of a shell is the distance from \( y \) to \( y = -2 \): \[ r(y) = y - (-2) = y + 2 \] - The **height** of the shell is the difference between the \( x \)-values on the right and left curves: \[ h(y) = (y^2 + 1) - 2y^2 = 1 - y^2 \] - The **volume element**: \[ dV = 2\pi \times r(y) \times h(y) \, dy = 2\pi (y + 2)(1 - y^2) dy \] - Limits of integration: \( y = -1 \) to \( y = 1 \). --- ### **Step 4: Write the integral** \[ V = 2\pi \int_{-1}^{1} (y + 2)(1 - y^2) \, dy \] --- ### **Step 5: Expand the integrand** \[ (y + 2)(1 - y^2) = y(1 - y^2) + 2(1 - y^2) = y - y^3 + 2 - 2y^2 \] So, \[ V = 2\pi \int_{-1}^{1} \left( y - y^3 + 2 - 2 y^2 \right) dy \] --- ### **Step 6: Integrate term by term** - \(\int y \, dy = \frac{1}{2} y^2\) - \(\int - y^3 \, dy = - \frac{1}{4} y^4\) - \(\int 2 \, dy = 2 y\) - \(\int -2 y^2 \, dy = - \frac{2}{3} y^3\) Thus, \[ V = 2\pi \left[ \frac{1}{2} y^2 - \frac{1}{4} y^4 + 2 y - \frac{2}{3} y^3 \right]_{-1}^{1} \] --- ### **Step 7: Evaluate at \( y = 1 \) and \( y = -1 \)** At \( y=1 \): \[ \frac{1}{2}(1)^2 - \frac{1}{4}(1)^4 + 2(1) - \frac{2}{3}(1)^3 = \frac{1}{2} - \frac{1}{4} + 2 - \frac{2}{3} \] At \( y=-1 \): \[ \frac{1}{2}(1)^2 - \frac{1}{4}(1)^4 + 2(-1) - \frac{2}{3}(-1)^3 = \frac{1}{2} - \frac{1}{4} - 2 + \frac{2}{3} \] Now, compute: At \( y=1 \): \[ \frac{1}{2} - \frac{1}{4} + 2 - \frac{2}{3} = \left( \frac{2}{4} - \frac{1}{4} \right) + 2 - \frac{2}{3} = \frac{1}{4} + 2 - \frac{2}{3} \] Express all with denominator 12: \[ \frac{3}{12} + \frac{24}{12} - \frac{8}{12} = \frac{3 + 24 - 8}{12} = \frac{19}{12} \] At \( y=-1 \): \[ \frac{1}{2} - \frac{1}{4} - 2 + \frac{2}{3} = \frac{1}{4} - 2 + \frac{2}{3} \] Express with denominator 12: \[ \frac{3}{12} - \frac{24}{12} + \frac{8}{12} = \frac{3 - 24 + 8}{12} = \frac{-13}{12} \] Subtract: \[ \frac{19}{12} - \left( - \frac{13}{12} \right) = \frac{19}{12} + \frac{13}{12} = \frac{32}{12} = \frac{8}{3} \] --- ### **Step 8: Final volume** \[ V = 2\pi \times \frac{8}{3} = \frac{16\pi}{3} \] --- ## **Final answer:** \[ \boxed{V = \frac{16\pi}{3}} \] --- ### **Summary:** - The region between \( x=2 y^2 \) and \( x= y^2 + 1 \) from \( y = -1 \) to \( y=1 \). - Rotated about \( y = -2 \). - Using the shell method with horizontal shells. - **Volume: \(\boxed{\frac{16\pi}{3}}\).** --- Let me know if you'd like further clarifications or additional problems!

Sure! Here are the final answers: **Problem 1:** \(\boxed{\frac{176\pi}{15}}\) **Problem 2:** \(\boxed{\frac{16\pi}{3}}\)