Find the surface area generated by rotating the first quadrant portion of the curve x\textsuperscript{2}=16-8y about the y-axis. ? 64.25 ? 66.38 ? 61.27 ? 58.41
The Correct Answer and Explanation is:
To find the surface area generated by rotating the first-quadrant portion of the curvex2=16−8yx^2 = 16 – 8yx2=16−8y
about the y-axis, we use the formula for surface area of revolution:A=2π∫y1y2x1+(dxdy)2 dyA = 2\pi \int_{y_1}^{y_2} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dyA=2π∫y1y2x1+(dydx)2dy
Step 1: Solve for xxx in terms of yyy
Given:x2=16−8y⇒x=16−8yx^2 = 16 – 8y \Rightarrow x = \sqrt{16 – 8y}x2=16−8y⇒x=16−8y
(Since we’re in the first quadrant, x≥0x \geq 0x≥0)
Step 2: Find bounds for yyy
The curve intersects the x-axis when y=0y = 0y=0:x2=16⇒x=4x^2 = 16 \Rightarrow x = 4×2=16⇒x=4
The curve intersects the y-axis when x=0x = 0x=0:0=16−8y⇒y=20 = \sqrt{16 – 8y} \Rightarrow y = 20=16−8y⇒y=2
So we integrate from y=0y = 0y=0 to y=2y = 2y=2.
Step 3: Find dxdy\frac{dx}{dy}dydx
x=(16−8y)1/2⇒dxdy=12(16−8y)−1/2⋅(−8)=−416−8yx = (16 – 8y)^{1/2} \Rightarrow \frac{dx}{dy} = \frac{1}{2}(16 – 8y)^{-1/2} \cdot (-8) = \frac{-4}{\sqrt{16 – 8y}}x=(16−8y)1/2⇒dydx=21(16−8y)−1/2⋅(−8)=16−8y−4
Step 4: Plug into surface area formula
A=2π∫0216−8y⋅1+(416−8y)2 dyA = 2\pi \int_0^2 \sqrt{16 – 8y} \cdot \sqrt{1 + \left(\frac{4}{\sqrt{16 – 8y}}\right)^2} \, dyA=2π∫0216−8y⋅1+(16−8y4)2dy
Simplify the expression inside the integral:(416−8y)2=1616−8y\left(\frac{4}{\sqrt{16 – 8y}}\right)^2 = \frac{16}{16 – 8y}(16−8y4)2=16−8y16
So:A=2π∫0216−8y⋅1+1616−8y dy=2π∫0216−8y⋅16−8y+1616−8y dyA = 2\pi \int_0^2 \sqrt{16 – 8y} \cdot \sqrt{1 + \frac{16}{16 – 8y}} \, dy = 2\pi \int_0^2 \sqrt{16 – 8y} \cdot \sqrt{\frac{16 – 8y + 16}{16 – 8y}} \, dyA=2π∫0216−8y⋅1+16−8y16dy=2π∫0216−8y⋅16−8y16−8y+16dy=2π∫0216−8y⋅32−8y16−8y dy= 2\pi \int_0^2 \sqrt{16 – 8y} \cdot \sqrt{\frac{32 – 8y}{16 – 8y}} \, dy=2π∫0216−8y⋅16−8y32−8ydy=2π∫0232−8y dy= 2\pi \int_0^2 \sqrt{32 – 8y} \, dy=2π∫0232−8ydy
Let’s substitute:
Let u=32−8y⇒du=−8dy⇒dy=−18duu = 32 – 8y \Rightarrow du = -8dy \Rightarrow dy = -\frac{1}{8}duu=32−8y⇒du=−8dy⇒dy=−81du
When y=0y = 0y=0, u=32u = 32u=32; when y=2y = 2y=2, u=16u = 16u=16A=2π∫3216u⋅(−18)du=π4∫1632u1/2duA = 2\pi \int_{32}^{16} \sqrt{u} \cdot \left(-\frac{1}{8}\right) du = \frac{\pi}{4} \int_{16}^{32} u^{1/2} duA=2π∫3216u⋅(−81)du=4π∫1632u1/2du
Now integrate:π4⋅[23u3/2]1632=π4⋅23(323/2−163/2)\frac{\pi}{4} \cdot \left[\frac{2}{3} u^{3/2} \right]_{16}^{32} = \frac{\pi}{4} \cdot \frac{2}{3} \left(32^{3/2} – 16^{3/2} \right)4π⋅[32u3/2]1632=4π⋅32(323/2−163/2)=π6(323−163)=π6((42)3−43)= \frac{\pi}{6} \left( \sqrt{32}^3 – \sqrt{16}^3 \right) = \frac{\pi}{6} \left( (4\sqrt{2})^3 – 4^3 \right)=6π(323−163)=6π((42)3−43)=π6(642−64)=64π6(2−1)= \frac{\pi}{6} \left( 64\sqrt{2} – 64 \right) = \frac{64\pi}{6} (\sqrt{2} – 1)=6π(642−64)=664π(2−1)=32π3(2−1)= \frac{32\pi}{3} (\sqrt{2} – 1)=332π(2−1)
Now approximate:2≈1.4142⇒(2−1)≈0.4142\sqrt{2} \approx 1.4142 \Rightarrow (\sqrt{2} – 1) \approx 0.41422≈1.4142⇒(2−1)≈0.4142A≈32π3⋅0.4142≈100.533≈33.51A \approx \frac{32\pi}{3} \cdot 0.4142 \approx \frac{100.53}{3} \approx 33.51A≈332π⋅0.4142≈3100.53≈33.51A≈33.51⋅π≈33.51⋅3.1416≈105.19A \approx 33.51 \cdot \pi \approx 33.51 \cdot 3.1416 \approx \boxed{105.19}A≈33.51⋅π≈33.51⋅3.1416≈105.19
Wait — we have an inconsistency. Let’s recheck:
Our earlier simplification was incorrect after taking the square root of that ratio.
Let’s back up and numerically evaluate the correct integral:
Correct Final Setup (for numerical integration):
A=2π∫0216−8y⋅1+1616−8y dy=2π∫0216−8y+16 dy=2π∫0232−8y dyA = 2\pi \int_0^2 \sqrt{16 – 8y} \cdot \sqrt{1 + \frac{16}{16 – 8y}} \, dy = 2\pi \int_0^2 \sqrt{16 – 8y + 16} \, dy = 2\pi \int_0^2 \sqrt{32 – 8y} \, dyA=2π∫0216−8y⋅1+16−8y16dy=2π∫0216−8y+16dy=2π∫0232−8ydy
Let’s integrate:∫0232−8y dy=−18⋅∫3216u du=−18⋅[23u3/2]3216=−18⋅23(163/2−323/2)=−18⋅23(64−181.02)=−18⋅23(−117.02)=18⋅23⋅117.02=234.0424=9.75Now:\[A=2π⋅9.75≈61.27\int_0^2 \sqrt{32 – 8y} \, dy = \frac{-1}{8} \cdot \int_{32}^{16} \sqrt{u} \, du = \frac{-1}{8} \cdot \left[ \frac{2}{3} u^{3/2} \right]_{32}^{16} = \frac{-1}{8} \cdot \frac{2}{3} (16^{3/2} – 32^{3/2}) = \frac{-1}{8} \cdot \frac{2}{3} (64 – 181.02) = \frac{-1}{8} \cdot \frac{2}{3} (-117.02) = \frac{1}{8} \cdot \frac{2}{3} \cdot 117.02 = \frac{234.04}{24} = 9.75 Now: \[ A = 2\pi \cdot 9.75 \approx 61.27∫0232−8ydy=8−1⋅∫3216udu=8−1⋅[32u3/2]3216=8−1⋅32(163/2−323/2)=8−1⋅32(64−181.02)=8−1⋅32(−117.02)=81⋅32⋅117.02=24234.04=9.75Now:\[A=2π⋅9.75≈61.27
✅ Final Answer: 61.27
📘 Textbook-style Explanation:
To find the surface area generated by rotating a curve around the y-axis, we use the formulaA=2π∫abx1+(dxdy)2 dyA = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dyA=2π∫abx1+(dydx)2dy
For the curve x2=16−8yx^2 = 16 – 8yx2=16−8y, we solve for xxx and get x=16−8yx = \sqrt{16 – 8y}x=16−8y.
Differentiating with respect to yyy, we find dxdy=−416−8y\frac{dx}{dy} = \frac{-4}{\sqrt{16 – 8y}}dydx=16−8y−4.
This leads to the surface area integral:A=2π∫0216−8y⋅1+1616−8y dy=2π∫0232−8y dyA = 2\pi \int_0^2 \sqrt{16 – 8y} \cdot \sqrt{1 + \frac{16}{16 – 8y}} \, dy = 2\pi \int_0^2 \sqrt{32 – 8y} \, dyA=2π∫0216−8y⋅1+16−8y16dy=2π∫0232−8ydy
With a substitution and evaluation, we find the surface area is approximately61.27 square units\boxed{61.27 \text{ square units}}61.27 square units
