- #1
BrownianMan
- 133
- 0
The base of each solid below is the region in the xy-plane bounded by the x-axis, the graph of \[y=\sqrt{x}\] and the line x = 3. Find the volume of each solid.
a) Each cross section is perpendicular to the y-axis is a square with one side in the xy-plane.
b) Each cross section is perpendicular to the y-axis is an isosceles right triangles with hypotenuse in the xy-plane.
This is what I have:
a)
\[V=\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{5}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{5}\]
b)
hyp=y^2
\[A=\frac{1}{2}\left ( \frac{y^2}{\sqrt{2}} \right )^2=\frac{y^4}{4}\]
\[V=\frac{1}{4}\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{20}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{20}\]
However, the answer in the back of the book is \[\frac{24\sqrt{3}}{5}\] for a) and \[\frac{6\sqrt{3}}{5}\] for b). What am I doing wrong?
a) Each cross section is perpendicular to the y-axis is a square with one side in the xy-plane.
b) Each cross section is perpendicular to the y-axis is an isosceles right triangles with hypotenuse in the xy-plane.
This is what I have:
a)
\[V=\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{5}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{5}\]
b)
hyp=y^2
\[A=\frac{1}{2}\left ( \frac{y^2}{\sqrt{2}} \right )^2=\frac{y^4}{4}\]
\[V=\frac{1}{4}\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{20}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{20}\]
However, the answer in the back of the book is \[\frac{24\sqrt{3}}{5}\] for a) and \[\frac{6\sqrt{3}}{5}\] for b). What am I doing wrong?
Last edited: