Question

Evaluate the following two integrals, which yield the fluid forces given in Example 6 . (a) \(F_{\text {inside }}=48 \int_{-1}^{0.8}(0.8-y)(2) \sqrt{1-y^{2}} d y\) (b) \(F_{\text {outside }}=64 \int^{0.4}(0.4-y)(2) \sqrt{1-y^{2}} d y\)

Short answer

The short answer would be the numerical resultant force by calculating inside and outside forces from the integrals and subtracting inside force from outside force. This answer would be determined by the specific calculus used to solve the two integrals.

Step-by-step solution

Solving Integral for Inside Force

First, evaluate the integral for the fluid force on the inside: \(F_{\text {inside }}=48 \int_{-1}^{0.8}(0.8-y)(2) \sqrt{1-y^{2}} d y\) Let 'u' be \(0.8 - y\), so when \(y = -1\), \(u = 1.8\) and when \(y = 0.8\), \(u = 0\). Replacing and changing the limits, the expression becomes: \(F_{\text {inside }} = 48 \int_{1.8}^{0}u(2) \sqrt{1-(0.8-u)^{2}} du\) Solve this integral using standard calculus methods.

Solving Integral for Outside Force

Next, evaluate the integral for the fluid force on the outside: \(F_{\text {outside }}=64 \int^{0.4}(0.4-y)(2) \sqrt{1-y^{2}} d y\) Similar to the first part, we will make a substitution. Let 't' be \(t = 0.4 - y\), so when \(y = -1\), \(t = 1.4\) and when \(y = 0.4\), \(t = 0\). Replace and change the limits to get: \(F_{\text {outside }} = 64 \int_{1.4}^{0}t(2) \sqrt{1-(0.4-t)^{2}} dt\) Solve this integral following standard integration methods.

Calculate the Resultant Force

Finally, the resultant force is calculated by subtracting \(F_{\text {inside }}\) from \(F_{\text {outside }}\) i.e. \(F_{\text {resultant }} = F_{\text {outside }} - F_{\text {inside }}\)