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Question
Math
Posted 6 months ago

Let RR be the region enclosed by y=34xy=3 \sqrt{4-x} and the axes in the first quadrant.

Region RR is the base of a solid. For each yy-value, the cross section of the solid taken perpendicular to the yy-axis is a rectangle whose base lies in RR and whose height is yy.

Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) 04(4y29)ydy\int_{0}^{4}\left(4-\frac{y^{2}}{9}\right) \cdot y d y
(B) 06(4y29)ydy\int_{0}^{6}\left(4-\frac{y^{2}}{9}\right) \cdot y d y
(c) 0434yydy\int_{0}^{4} 3 \sqrt{4-y} \cdot y d y
() 0634yydy\int_{0}^{6} 3 \sqrt{4-y} \cdot y d y
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Answer from Sia
Posted 6 months ago
Solution by Steps
step 2
The cross-section of the solid perpendicular to the y-axis is a rectangle with base in RR and height yy
step 3
The base of the rectangle at a given yy is the horizontal distance from the y-axis to the curve, which is x=4y29x = 4 - \frac{y^2}{9}
step 4
The volume of the solid is given by the integral of the area of the cross-section along the y-axis. The area of the cross-section is ybase=y(4y29)y \cdot \text{base} = y \cdot (4 - \frac{y^2}{9})
step 5
The correct integral to find the volume is 0634yydy\int_{0}^{6} 3\sqrt{4-y} \cdot y \, dy
D
Key Concept
Volume of a solid with cross-sectional area
Explanation
The volume of a solid can be found by integrating the area of its cross-sections along the axis perpendicular to the cross-sections.

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