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BrownianMan
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The base of each solid below is the region in the xy-plane bounded by the x-axis, the graph of [tex]\[y=\sqrt{x}\][/tex] 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)
[tex]\[V=\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{5}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{5}\][/tex]
b)
[tex]hyp=y^2[/tex]
[tex]\[A=\frac{1}{2}\left ( \frac{y^2}{\sqrt{2}} \right )^2=\frac{y^4}{4}\][/tex]
[tex]\[V=\frac{1}{4}\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{20}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{20}\][/tex]
However, the answer in the back of the book is [tex]\[\frac{24\sqrt{3}}{5}\][/tex] for a) and [tex]\[\frac{6\sqrt{3}}{5}\][/tex] 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)
[tex]\[V=\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{5}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{5}\][/tex]
b)
[tex]hyp=y^2[/tex]
[tex]\[A=\frac{1}{2}\left ( \frac{y^2}{\sqrt{2}} \right )^2=\frac{y^4}{4}\][/tex]
[tex]\[V=\frac{1}{4}\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{20}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{20}\][/tex]
However, the answer in the back of the book is [tex]\[\frac{24\sqrt{3}}{5}\][/tex] for a) and [tex]\[\frac{6\sqrt{3}}{5}\][/tex] for b). What am I doing wrong?
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