Question

Use the general slicing method to find the volume of the solid whose base is the triangle with vertices (0,0), (7,0), and (0,7) and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles.
(need exact answer in terms of pi).

Answer 1

Answer:

The answer is "$\bold{\frac{343 \ \pi}{24} \ \text{cubic units}}$"

Step-by-step explanation:

The volume of the mass whose cross-sectional area has been perpendicular is usually sliced by the method The cross-section, which is based to parallel to y-axis;  

$V = \int_{a}^{b} A (x)dx$

The semi-circular segment of the strong and seems to be perpendicular to foundation and  

Y-axis parallel.  

Its cross-section does have a diameter of: $(7-x)$.  

It also transverse radius is: $\frac{1}{2}(7-x)$.  

The semi-circular segment area is,

$Formula: \\\\ A(x)=\frac{1}{2} \times \pi \times r^2$

        $=\frac{1}{2} \times \pi \times (\frac{(7-x)}{2})^2\\\\=\frac{1}{2} \times \pi \times (\frac{(7-x)^2}{4})\\\\=\frac{1}{8}\pi(7-x)^2$

when

$0 \leq \ x \ \leq 7$

Calculating the volume from the solid accordingly:

$V= \int_{0}^{7}\frac{1}{8} \times\pi \times (7-x)^2 dx$

   $= \frac{1}{8} \pi \int_{0}^{7}(7-x)^2 dx \\\\= \frac{1}{8} \pi [\frac{(7-3)^3}{-3}]_{0}^{7}\\\\= \frac{\pi}{24} \times [7^3-0]\\\\= \frac{\pi}{24} \times 343\\\\= \frac{343 \pi}{24}$