Problem

Find the surface area of the surface generated by revolving \( y=e^{2 x} \) with \( 0 \leq x \leq 1 \) about the \( x \)-axis.

Answer

Expert–verified
Hide Steps
Answer

\int_{0}^{1} 2\pi e^{2 x} \sqrt{1+4e^{4 x}} dx

Steps

Step 1 ：\int_{0}^{1} 2\pi y \sqrt{1+{\left( \frac{dy}{dx} \right)}^2} dx

Step 2 ：\frac{dy}{dx} = 2e^{2 x}

Step 3 ：1 + {\left(\frac{dy}{dx}\right)}^2 = 1 + 4e^{4 x}

Step 4 ：\int_{0}^{1} 2\pi e^{2 x} \sqrt{1+4e^{4 x}} dx

link_gpt

Popular Problems

Compute $\int_{0}^{\ln (2) / … Solve the triangle. \[ \mathr… QUESTION 2 - 1 POINT Solve th… 5. Suppose you are a computer… Answer the statistical measur… $(-\infty, 2) \cap(-6, \infty… A firm does not pay a dividen… Math 112 Section: 4. Tubercul… Find $y^{\prime}$ if $y=\ln \… 3. (3pts) Cherry trees in a c… More