The solid lies between planes perpendicular to the x axis at \(\displaystyle x = 0\) and \(\displaystyle x = 4\). The cross-sections perpendicular to the x axis on the interval \(\displaystyle 0 \leq x \leq 4\) are squares whose diagonals run from the parabola \(\displaystyle y = -\sqrt{x}\) to the parabola \(\displaystyle y = \sqrt{x}\). Find the volume
\(\displaystyle V = \int_{a}^{b} A(x) dx\)
\(\displaystyle A = [\dfrac{2 \sqrt{x}}{2}]^{2}\)
\(\displaystyle A = \dfrac{4x}{4}\)
\(\displaystyle A = x\)
\(\displaystyle V = \int_{0}^{4} x dx\)
\(\displaystyle V = \dfrac{x^{2}}{2} \)
\(\displaystyle V = \dfrac{(4)^{2}}{2} - 0\)
\(\displaystyle V = \dfrac{16}{2} = 8\) Again, answer in book is 16. Where am I going wrong here?
How about this way?
\(\displaystyle V = \int_{a}^{b} A(x) dx\)
\(\displaystyle A = [\dfrac{2 \sqrt{x}}{2}]\)
\(\displaystyle A = \dfrac{2\sqrt{x}}{2}\)
\(\displaystyle A = \sqrt{x}\)
\(\displaystyle V = \int_{0}^{4} \sqrt{x} dx\)
\(\displaystyle V = \int_{0}^{4} x^{1/2} dx\)
\(\displaystyle V = \dfrac{x^{3/2}}{\dfrac{3}{2}}\)
\(\displaystyle V = \dfrac{2}{3}x^{3/2}\)
\(\displaystyle V = \dfrac{2}{3}(4)^{3/2} - 0 = \dfrac{16}{3}\) Also wrong
Finally, how about this way?
The solid lies between planes perpendicular to the x axis at \(\displaystyle x = 0\) and \(\displaystyle x = 4\). The cross-sections perpendicular to the x axis on the interval \(\displaystyle 0 \leq x \leq 4\) are squares whose diagonals run from the parabola \(\displaystyle y = -\sqrt{x}\) to the parabola \(\displaystyle y = \sqrt{x}\). Find the volume
\(\displaystyle V = \int_{a}^{b} A(x) dx\)
\(\displaystyle x^{2}\) is the area of a square
\(\displaystyle A = x^{2}\)
\(\displaystyle V = \int_{0}^{4}[x] [2 \sqrt{x}]^{2} dx\)
\(\displaystyle V = \int_{0}^{4} [x][4x] dx\)
\(\displaystyle V = \dfrac{4x^{2}}{2} \)
\(\displaystyle V = 2x^{2} \)
\(\displaystyle V = 2(4)^{2} - 0 = 32\)
OK, this way looks better. But still not 16. Maybe the diagonal thing needs to be put in.
