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)

please explain. The answer I was getting was 49pi/8, but my homework says that is wrong.

Answer 1

Figure is attached below:

Answer:

$100.04\pi$

Step-by-step explanation:

Given Data:

triangle vertices= A,B,C=(0,0), (7,0), (0,7)

Required: Find Volume of the solid=V=?

For this type of question we need to use the following formula of slicing method

$\int\limits^a_bA {(x)} \, dx................Eq(1)$

where a and b are limits which shows that solid extends from $x=a\ to\ x=b$ with the known cross section area  $A(x)$  perpendicular to x-axis.

From the Figure.2 we can see that limits are from $x=0$ to $x=7$ and the red line is the diameter of the given semi circle.

We can see that if x increases then the diameter decreases along the hypotenuse thus diameter of the semi circle$=d=$ $7-x$

Radius is half of the diameter, so Radius of semi circle$=r=\frac{7-x}{2}$

Here, Area $A(x)$ is the area of the semi circle

so $A(x)=\frac{1}{2}(\pi ) r^{2}$

Putting values, we get

$A(x)=\frac{1}{2}(\pi ) (\frac{7-x}{2}) ^{2}$

Putting values in Eq(1), we get

$V=\int\limits^7_0 {\frac{1}{2}(\pi ) (\frac{7-x}{2}) ^{2}} \, dx$

$V=\frac{\pi }{2}\int\limits^0_7 {(\frac{7-x}{2} )^{2} } \, dx$

$V=\frac{\pi }{2}\int\limits^0_7 {(\frac{49+x^{2} -2(7)(x)}{4} ) } \, dx$ expanding formula ($(a+b)^{2}=a^{2}+b^{2}+2ab$)

$V=\frac{\pi }{8}\int\limits^7_0 ({49+x^{2} -14x)} \, dx$

$V=\frac{\pi }{8}(\int\limits^7_0 {49} \, dx+\int\limits^7_0 {x^{2} } \, dx +\int\limits^7_0 {14x} \, dx )$

$V=\frac{\pi }{8}(49x+\frac{x^3}{3}+\frac{14x^2}{2}) (for\ x=0\ to\ 7)$

Putting $x=0$ we get everything zero that's why only have expression for value of $x=7$ as:

$V=\frac{\pi }{8}(49*7+\frac{7^{3}}{3}+14*\frac{7^2}{2} )$

$V=\frac{\pi }{8}(343+114.33+343)$

$V=\frac{\pi }{8}(800.33)\\\\ V=100.04\pi }$