Find the area of the part of the surface $y = 8 x + z ^ { 2 }$ that lies between the planes $x = 0 , x = 4 , z = 0$, and $z = 1$. Select the correct answer.
A) $32.5768$
B) $39.7002$
C) $44.7002$
D) $41.7002$
E) $42.7002$

Question

Quizplus · Accepted Answer

The surface area is given by the integral of the square root of $1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2$ over the region. Here, $\frac{\partial z}{\partial x} = 0$ and $\frac{\partial z}{\partial y} = 16 x$, so the surface area is$$\int_0^4 \int_0^1 \sqrt{1 + (16x)^2} \ dz \ dx = \int_0^4 \left[ z\sqrt{1 + (16x)^2} \right]_0^1 \ dx = \int_0^4 \sqrt{1 + (16x)^2} \ dx$$This integral can be evaluated using a trigonometric substitution, $x = \frac{1}{16} \tan \theta$. Then, $dx = \frac{1}{16} \sec^2 \theta \ d\theta$, and$$\begin{split} \int_0^4 \sqrt{1 + (16x)^2} \ dx &= \int_0^{\tan^{-1} (64)} \sqrt{1 + \tan^2 \theta} \ \frac{1}{16} \sec^2 \theta \ d\theta \\ &= \int_0^{\tan^{-1} (64)} \sec^3 \theta \ d\theta \\ &= \left[ \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| \right]_0^{\tan^{-1} (64)} \\ &= \frac{1}{2} (64\sqrt{1 + 64^2} + \ln |64 + \sqrt{1 + 64^2}|) \\ &\approx 32.5768 \end{split}$$

Find the Area of the Part of the Surface y=8x+z2y = 8 x + z ^ { 2 }y=8x+z2

Find the Area of the Part of the Surface $y = 8 x + z ^ { 2 }$