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Give step-by-step solution with explanation and final answer:" 25 1 Cylinders and Quadric Surfaces (13.6) a 1. Come te fin cmon of unc tc p orem (0) in ht des ek pps nh cin, she Kd oe Se va pe of hit: Chale or we sd bl ly ut oe apg so ht (Sn oe 30 gh of he cy etching vo 20 fe eto ses a ie yt igs i eae ox ht Ne

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Give step-by-step solution with explanation and final answer:Uploaded Image" 25 1 Cylinders and Quadric Surfaces (13.6) a 1. Come te fin cmon of unc tc p orem (0) in ht des ek pps nh cin, she Kd oe Se va pe of hit: Chale or we sd bl ly ut oe apg so ht (Sn oe 30 gh of he cy etching vo 20 fe eto ses a ie yt igs i eae ox ht Ne

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Created at: 2025-05-19 09:50:17

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Let's solve the problem step by step: ## **Given Equation:** \[ 9y^2 + 4x^2 = 36 \] --- ### **(a) What kind of surface is this?** **Step 1:** Notice that **z does not appear** in the equation. - This means the equation is independent of z, so the surface is the same for every value of \( z \). - Such a surface is called a **cylinder** (specifically, a "cylindrical surface" since it extends infinitely along the \( z \)-axis). **Step 2:** The equation \( 9y^2 + 4x^2 = 36 \) is an **ellipse** in the \( xy \)-plane. - Since it extends infinitely in the \( z \) direction, the surface is an **elliptic cylinder**. **Final Answer (a):** > The surface is an **elliptic cylinder**. --- ### **(b) Sketch the \( yx \)-trace of this surface.** The \( yx \)-trace is just the curve \( 9y^2 + 4x^2 = 36 \) in the \( xy \)-plane. **Step 1:** Rewrite in standard form: \[ \frac{y^2}{4} + \frac{x^2}{9} = 1 \] **Step 2:** Identify axes intercepts: - For \( y \)-intercept (\( x = 0 \)): \( \frac{y^2}{4} = 1 \implies y^2 = 4 \implies y = \pm 2 \) - For \( x \)-intercept (\( y = 0 \)): \( \frac{x^2}{9} = 1 \implies x^2 = 9 \implies x = \pm 3 \) **Step 3:** Sketch - Draw the \( x \)-axis and \( y \)-axis. - Mark intercepts at \( (3, 0) \), \( (-3, 0) \), \( (0, 2) \), \( (0, -2) \). - Draw an ellipse passing through these points. **(Sketch example below):** ``` y ^ 2 | *----* | / \ | / \ | / \ -2 |*------------------* +---------------------> x -3 0 3 ``` **Final Answer (b):** > The \( yx \)-trace is an ellipse centered at the origin, with \( x \)-intercepts at \( x = \pm 3 \) and \( y \)-intercepts at \( y = \pm 2 \). --- ### **(c) Sketch a 3D graph of the surface** **Step 1:** Extend the 2D ellipse in the \( z \)-direction. - For every value of \( z \), the cross-section in the \( xy \)-plane is the same ellipse: \( 9y^2 + 4x^2 = 36 \). **Step 2:** Draw axes \( x, y, z \). - Draw the ellipse \( 9y^2 + 4x^2 = 36 \) in the \( xy \)-plane. - Show this ellipse is present for every \( z \) value by drawing several ellipses stacked in the \( z \)-direction, forming a "tube" or cylinder. **(Sketch example below):** ``` z | | ______ | / / | / / | / / | /______/ +------------------- y / / x (Imagine the elliptical shape extending infinitely along the z-axis) ``` **Final Answer (c):** > The surface is an elliptic cylinder with cross-sections \( 9y^2 + 4x^2 = 36 \) in the \( xy \)-plane, extending infinitely along the \( z \)-axis. The ellipse has \( x \)-intercepts at \( x = \pm 3 \), \( y \)-intercepts at \( y = \pm 2 \), for every value of \( z \). --- **Summary Table:** | Part | Answer | |------|--------| | (a) | Elliptic cylinder | | (b) | Ellipse: \( x = \pm 3 \), \( y = \pm 2 \) | | (c) | Elliptic cylinder in 3D, ellipse in \( xy \)-plane, extends along \( z \) | Let me know if you need a more detailed hand-drawn sketch!

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