MHT CET · Maths · Area Under Curves
The volume of solid generated by revolving about the \(y\) -axis the figure bounded by the parabola \(y=x^{2}\) and \(x=y^{2}\) is
- A \(\frac{21}{5} \pi\)
- B \(\frac{24}{5} \pi\)
- C \(\frac{2}{15} \pi\)
- D \(\frac{5}{24} \pi\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{15} \pi\)
Step-by-step Solution
Detailed explanation
\(V=\int_{0}^{1} \pi x^{2} d y\)
\(=\left|\pi \int_{0}^{1}\left(y^{4}-y\right) d y\right|\)
\(=\left|\pi\left[\frac{y^{5}}{5}-\frac{y^{2}}{2}\right]_{0}^{1}\right|\)
\(=\left|\pi\left[\frac{1}{5}-\frac{1}{3}\right]\right|\)
\(=\left|\frac{-2 \pi}{15}\right|=\frac{2 \pi}{15}\)
\(=\left|\pi \int_{0}^{1}\left(y^{4}-y\right) d y\right|\)
\(=\left|\pi\left[\frac{y^{5}}{5}-\frac{y^{2}}{2}\right]_{0}^{1}\right|\)
\(=\left|\pi\left[\frac{1}{5}-\frac{1}{3}\right]\right|\)
\(=\left|\frac{-2 \pi}{15}\right|=\frac{2 \pi}{15}\)
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