Let's solve the **first problem step-by-step**: ### **Problem 1** **Given:** - Curves: \( y = \sin x \), \( y = \cos x \) - Interval: \( \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \) - Axis of rotation: \( y = -1 \) We are to find the volume of the solid formed by rotating the region bounded by these curves, in the specified interval, about the line \( y = -1 \). --- ## **Step 1: Sketch the Region** 1. **Graph \( y = \sin x \) and \( y = \cos x \) from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \):** - At \( x = \frac{\pi}{4} \), \( \sin x = \cos x = \frac{1}{\sqrt{2}} \). - At \( x = \frac{\pi}{2} \), \( \sin x = 1 \), \( \cos x = 0 \). 2. **The region is bounded above by \( y = \sin x \) and below by \( y = \cos x \), from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \).** --- ## **Step 2: Set Up the Volume Integral (Washer Method)** Since we are rotating around \( y = -1 \): - **Outer radius** (from \( y = -1 \) to \( y = \sin x \)): \( R_{\text{outer}} = \sin x - (-1) = \sin x + 1 \) - **Inner radius** (from \( y = -1 \) to \( y = \cos x \)): \( R_{\text{inner}} = \cos x - (-1) = \cos x + 1 \) Area of a washer at \( x \): \[ A(x) = \pi \left[ (\sin x + 1)^2 - (\cos x + 1)^2 \right] \] --- ## **Step 3: Write the Volume Integral** \[ V = \int_{x = \frac{\pi}{4}}^{x = \frac{\pi}{2}} \pi \left[ (\sin x + 1)^2 - (\cos x + 1)^2 \right] dx \] --- ## **Step 4: Simplify the Integrand** \[ (\sin x + 1)^2 = \sin^2 x + 2\sin x + 1 \] \[ (\cos x + 1)^2 = \cos^2 x + 2\cos x + 1 \] So, \[ (\sin x + 1)^2 - (\cos x + 1)^2 = (\sin^2 x - \cos^2 x) + 2(\sin x - \cos x) \] Recall: \( \sin^2 x - \cos^2 x = -\cos(2x) \) So, \[ = -\cos(2x) + 2(\sin x - \cos x) \] --- ## **Step 5: Integrate** So, \[ V = \pi \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left[ -\cos(2x) + 2(\sin x - \cos x) \right] dx \] Integrate term by term: - \( \int -\cos(2x) dx = -\frac{1}{2}\sin(2x) \) - \( \int 2\sin x dx = -2\cos x \) - \( \int -2\cos x dx = -2\sin x \) So, \[ V = \pi \left[ -\frac{1}{2}\sin(2x) - 2\cos x - 2\sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \] --- ## **Step 6: Plug in the Bounds** **At \( x = \frac{\pi}{2} \):** - \( \sin(2x) = \sin(\pi) = 0 \) - \( \cos x = \cos(\frac{\pi}{2}) = 0 \) - \( \sin x = \sin(\frac{\pi}{2}) = 1 \) So, \[ F\left(\frac{\pi}{2}\right) = 0 - 0 - 2(1) = -2 \] **At \( x = \frac{\pi}{4} \):** - \( \sin(2x) = \sin(\frac{\pi}{2}) = 1 \) - \( \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \) - \( \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \) So, \[ F\left(\frac{\pi}{4}\right) = -\frac{1}{2}(1) - 2\left(\frac{1}{\sqrt{2}}\right) - 2\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2} - \frac{4}{\sqrt{2}} = -\frac{1}{2} - 2\sqrt{2} \] --- ## **Step 7: Compute the Difference** \[ V = \pi \left[ (-2) - \left(-\frac{1}{2} - 2\sqrt{2}\right) \right] = \pi \left( -2 + \frac{1}{2} + 2\sqrt{2} \right) = \pi \left( -\frac{3}{2} + 2\sqrt{2} \right) \] --- ## **Final Answer** \[ \boxed{ V = \pi \left( 2\sqrt{2} - \frac{3}{2} \right) } \] --- If you need a sketch, let me know!

Certainly! It appears that the second part involves a different region and axis of rotation. Based on the provided image and description, I'll interpret and proceed with the second problem. --- ## **Problem 2** **Given:** - The region bounded by \( y = x^2 \) and \( y = 4 \) - Rotation about \( x = 2 \) Our goal is to find the volume of the solid obtained by rotating this region about the vertical line \( x = 2 \). --- ## **Step 1: Sketch the Region** - The parabola \( y = x^2 \) intersects \( y = 4 \) at \( x = \pm 2 \). - The bounded region is between \( x = -2 \) and \( x = 2 \), from \( y = x^2 \) up to \( y=4 \). --- ## **Step 2: Identify the Axis of Rotation** - The axis is the vertical line \( x = 2 \). - Since the region spans from \( x = -2 \) to \( x = 2 \), and the axis is at \( x=2 \), the region is to the left of the axis. --- ## **Step 3: Decide on the Method** - Rotation about a vertical line suggests using the **washer method** with respect to \( x \), or the **cylindrical shell method**. - **Shell method** is more straightforward here because: - Shell radius: distance from \( x \) to \( x=2 \) is \( |2 - x| \). - Shell height: \( y_{\text{top}} - y_{\text{bottom}} = 4 - x^2 \). --- ## **Step 4: Set Up the Shell Method Integral** - For \( x \) from \( -2 \) to \( 2 \): \[ V = 2\pi \int_{x=-2}^{x=2} \text{(radius)} \times \text{(height)} \, dx \] - Radius: \( R(x) = |2 - x| \). Since \( x \in [-2, 2] \), \( 2 - x \geq 0 \), so: \[ R(x) = 2 - x \] - Height: \( h(x) = 4 - x^2 \) Thus, \[ V = 2\pi \int_{-2}^{2} (2 - x)(4 - x^2) \, dx \] --- ## **Step 5: Simplify the Integrand** \[ (2 - x)(4 - x^2) = 2(4 - x^2) - x(4 - x^2) = 8 - 2x^2 - 4x + x^3 \] So, \[ V = 2\pi \int_{-2}^{2} (8 - 2x^2 - 4x + x^3) dx \] --- ## **Step 6: Integrate Term-by-Term** \[ \int 8 \, dx = 8x \] \[ \int -2x^2 \, dx = -2 \times \frac{x^3}{3} = -\frac{2}{3} x^3 \] \[ \int -4x \, dx = -2x^2 \] \[ \int x^3 \, dx = \frac{x^4}{4} \] Putting it all together: \[ V = 2\pi \left[ 8x - \frac{2}{3}x^3 - 2x^2 + \frac{x^4}{4} \right]_{-2}^{2} \] --- ## **Step 7: Evaluate at the bounds** **At \( x=2 \):** \[ 8(2) - \frac{2}{3}(8) - 2(4) + \frac{16}{4} = 16 - \frac{16}{3} - 8 + 4 \] Simplify: \[ 16 - 8 + 4 = 12 \] \[ \text{Subtract } \frac{16}{3} \Rightarrow 12 - \frac{16}{3} = \frac{36}{3} - \frac{16}{3} = \frac{20}{3} \] **At \( x=-2 \):** \[ 8(-2) - \frac{2}{3}(-8) - 2(4) + \frac{16}{4} = -16 + \frac{16}{3} - 8 + 4 \] Simplify: \[ -16 - 8 + 4 = -20 \] \[ Add \frac{16}{3} \Rightarrow -20 + \frac{16}{3} = -\frac{60}{3} + \frac{16}{3} = -\frac{44}{3} \] --- ## **Step 8: Compute the difference** \[ V = 2\pi \left( \frac{20}{3} - \left( -\frac{44}{3} \right) \right) = 2\pi \left( \frac{20}{3} + \frac{44}{3} \right) = 2\pi \times \frac{64}{3} = \frac{128\pi}{3} \] --- ## **Final Answer:** \[ \boxed{ V = \frac{128\pi}{3} } \] --- ## **Summary:** - **First problem:** The volume is \( \pi \left( 2\sqrt{2} - \frac{3}{2} \right) \). - **Second problem:** The volume is \( \frac{128\pi}{3} \). Let me know if you'd like sketches or further explanations!

Let's carefully analyze and solve the problem where the region bounded by \( x = y^2 \) and \( x = 1 - y^2 \) is revolved about the line \( x = 2 \). --- ## **Part (b): Find the volume of the solid obtained by rotating the region bounded by \( x = y^2 \) and \( x = 1 - y^2 \) about the line \( x=2 \).** --- ### **Step 1: Understand and sketch the region** - The curves are: - \( x = y^2 \) (a right-opening parabola) - \( x = 1 - y^2 \) (a left-opening parabola) - Find their intersection points: Set \( y^2 = 1 - y^2 \): \[ 2 y^2 = 1 \Rightarrow y^2 = \frac{1}{2} \Rightarrow y = \pm \frac{1}{\sqrt{2}} \] - The region is between \( y = -\frac{1}{\sqrt{2}} \) and \( y= \frac{1}{\sqrt{2}} \), bounded by these two curves. --- ### **Step 2: Describe the region** - For each \( y \) in \([-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\): - Left boundary: \( x = y^2 \) - Right boundary: \( x = 1 - y^2 \) - The region is symmetric about the \( y \)-axis. --- ### **Step 3: Choice of method** - Since rotation is about the vertical line \( x=2 \), and the region is bounded in \( x \), the **washer method** with respect to \( y \) is suitable. - The **radius** of a washer at a given \( y \) is the distance from the axis \( x=2 \) to the inner and outer edges of the region. --- ### **Step 4: Determine the radii** - For a point at \( (x, y) \): - Distance from \( x=2 \) is \( R = 2 - x \), since the region is to the left of \( x=2 \). - **Outer radius:** the maximum \( x \)-value at \( y \) is \( x = 1 - y^2 \). Since this is to the left of \( x=2 \): \[ R_{\text{outer}} = 2 - y^2 \] - **Inner radius:** the minimum \( x \)-value at \( y \) is \( x = y^2 \): \[ R_{\text{inner}} = 2 - y^2 \] Wait, but both are the same! That suggests the region is between the two curves, and when revolving around \( x=2 \), the **distance from the axis** to each boundary is: \[ \text{Inner radius} = 2 - (1 - y^2) = 2 - 1 + y^2 = 1 + y^2 \] \[ \text{Outer radius} = 2 - y^2 \] Alternatively, since \( x \) runs from \( y^2 \) (left boundary) to \( 1 - y^2 \) (right boundary), the distances from the axis \( x=2 \) are: - To \( x = y^2 \): \( R_{\text{inner}} = 2 - y^2 \) - To \( x= 1 - y^2 \): \( R_{\text{outer}} = 2 - (1 - y^2) = 1 + y^2 \) Note: The larger radius corresponds to the boundary closer to the axis \( x=2 \). Since \( y^2 \geq 0 \): - \( 2 - y^2 \) (outer boundary) - \( 1 + y^2 \) (inner boundary) Therefore, when revolving, the washers have: - Inner radius: \( R_{\text{inner}} = 1 + y^2 \) - Outer radius: \( R_{\text{outer}} = 2 - y^2 \) --- ### **Step 5: Set up the volume integral** The volume: \[ V = \pi \int_{y = -\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left[ R_{\text{outer}}^2 - R_{\text{inner}}^2 \right] dy \] Plugging in the radii: \[ V = \pi \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left[ (2 - y^2)^2 - (1 + y^2)^2 \right] dy \] --- ### **Step 6: Simplify the integrand** Calculate each square: \[ (2 - y^2)^2 = 4 - 4 y^2 + y^4 \] \[ (1 + y^2)^2 = 1 + 2 y^2 + y^4 \] Subtract: \[ (4 - 4 y^2 + y^4) - (1 + 2 y^2 + y^4) = (4 - 1) + (-4 y^2 - 2 y^2) + (y^4 - y^4) \] \[ = 3 - 6 y^2 + 0 = 3 - 6 y^2 \] --- ### **Step 7: Write the volume integral** \[ V = \pi \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (3 - 6 y^2) dy \] Since the integrand is an even function (\( y^2 \) symmetric), and the limits are symmetric, we can write: \[ V = 2 \pi \int_0^{\frac{1}{\sqrt{2}}} (3 - 6 y^2) dy \] --- ### **Step 8: Integrate** \[ \int (3 - 6 y^2) dy = 3 y - 2 y^3 \] Evaluate from 0 to \( y = \frac{1}{\sqrt{2}} \): \[ 3 \left( \frac{1}{\sqrt{2}} \right) - 2 \left( \frac{1}{\sqrt{2}} \right)^3 \] Compute: \[ 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \] \[ \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{1}{(\sqrt{2})^3} = \frac{1}{2 \sqrt{2}} \] So, \[ - 2 \times \frac{1}{2 \sqrt{2}} = - \frac{1}{\sqrt{2}} \] Putting it together: \[ \frac{3}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \] Multiply by 2 (from the symmetry): \[ V = 2 \pi \times \sqrt{2} = 2 \sqrt{2} \pi \] --- ## **Final Answer:** \[ \boxed{ V = 2 \sqrt{2} \pi } \] --- ## **Summary:** - The volume of the solid formed by revolving the region bounded by \( x = y^2 \) and \( x = 1 - y^2 \) about \( x=2 \) is **\( 2 \sqrt{2} \pi \)**. Let me know if you'd like further clarifications or sketches!