Home / AP Calculus BC : 8.9 Volume with Disc Method: Revolving  Around the x- or y-Axis- Exam Style questions with Answer- MCQ

AP Calculus BC : 8.9 Volume with Disc Method: Revolving  Around the x- or y-Axis- Exam Style questions with Answer- MCQ

Question

The region in the first quadrant bounded by the graph of\( y=secx,x=\frac{\pi }{4}\) , and the axes is rotated about the x-axis. What is the volume of the solid generated?

(A)\(\frac{\pi ^{2}}{4}\)                     (B)\(\pi -1 \)                                   (C)\(\pi\)                           (D)\(2\pi\)                                (E)\(\frac{8\pi }{3}\)

Answer/Explanation

 

Question 

The region R in the first quadrant is enclosed by the lines x = 0 and y = 5 and the graph of \(y= x^{2}+1\). The volume of the solid generated when R is revolved about the y-axis is
(A) 6π                                        (B) 8π                                                  (C) \(\frac{34π}{3}\)                                (D) 16π                                                 (E) \frac{544π}{15}\)

Answer/Explanation

Ans:B

Question

 When the region enclosed by the graphs of y = x and  \(y=4x-x^{2}\)  is revolved about the y-axis, the volume of the solid generated is given by

(A)\(\pi\int_{0}^{3}(x^3-3x^2)dx\)

(B)\(\pi\int_{0}^{3}\left ( x^2\left ( 4x-x^2 \right ) \right )dx\)

(C)\(\pi\int_{0}^{3}(3x-x^2)^{2}dx\)

(D)\(2\pi\int_{0}^{3}(x^3-3x^2)dx\)

(E)\(2\pi\int_{0}^{3}(3x^3-x^2)dx\)

Answer/Explanation

Ans:E

Question

 The base of a solid is the region in the first quadrant enclosed by the graph of\( y=2-x^2\) and the coordinate axes. If every cross section of the solid perpendicular to the y-axis is a square, the
volume of the solid is given by

(A)\(\pi\int_{0}^{2}(2-y)^{2}dy\)

(B)\(\int_{0}^{2}(2-y)^{2}dy\)

(C)\(\pi\int_{0}^{\sqrt{2}}\left ( 2-x^{2} \right )^{2}dx\)

(D)\(\int_{0}^{\sqrt{2}}\left ( 2-x^{2} \right )^{2}dx\)

(E)\(\int_{0}^{\sqrt{2}}\left ( 2-x^{2} \right )dx\)

Answer/Explanation

Ans:B

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