Volume by cylindrical shells of a rotated region

[MarkFL]

Here is the question:

Volume of solid (calc 2)?


Find the volume V of the solid whose base is the circle
x^2 + y^2 = 25
and whose cross sections perpendicular to the x-axis are triangles whose height and base are equal.

help appreciated

thanks

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello nick,

For an arbitrary slice of the described solid, the base of this triangular slice will be from the $y$-coordinate of the upper half to the $y$-coordinate of the lower half, or:

$$b=y-(-y)=2y$$

And thus, since the base and height are the same, and using the formula for the area of a triangle, we find the volume of the slice is:

$$dV=\frac{1}{2}(2y)(2y)\,dx=2y^2\,dx$$

Now, using the boundary of the base of the solid, we find:

$$2y^2=2\left(25-x^2 \right)$$

And so we obtain:

$$dV=2\left(25-x^2 \right)\,dx$$

Now, summing up the slices, we get:

$$V=2\int_{-5}^{5}25-x^2\,dx$$

And using the even-function rule, we may write:

$$V=4\int_{0}^{5}25-x^2\,dx$$

Applying the FTOC, there results:

$$V=4\left[25x-\frac{1}{3}x^3 \right]_{0}^{5}=4\cdot5^3\left(1-\frac{1}{3} \right)=\frac{(2\cdot5)^3}{3}=\frac{1000}{3}$$